Chemistry Subjective Questions and Solutions Module Paper 2
Illustration 1: A cylinder of gas is assumed to contain 11.2 kg butane. If a normal family needs 20,000 kJ of energy per day for cooking. How long will the cylinder last if the enthalpy of combustion DH = –2658 kJ for butane.
Solution: C4H10(g) + O2(g) ¾® 4CO2(g) + 5H2O(g)
DH = – 2658 kJ
Molecular weight of C4H10 = 58 g/mole
58 g of butane on combustion produces 2658 kJ heat
\ 11.2 kg of butane on combustion produces
Family needs 20,000 kJ of energy per day
\ No. of days = = 25.66 days
Illustration 2: Calculate heat of neutralisation from following data.
200 ml of 1 (M) HCl is mixed with 400 ml 0.5 (M) NaOH the temperature rise in calorimeter was found to be 4.4°C. Water equivalent of calormeter is 12g and specific heat is 1 cal ml–1 degree–1 for solution
Solution: Meq. of acid & base = 200
i.e. 200 Meq. of HCl react with 200 Meq. of NaOH to produce heat = DH
\ 1000 Meq. of HCl when react with 1000 Meq. of NaOH will give heat
= 5 ´ DH = heat of neutralisation
Now heat produced during neutralisation of 200 Meq. of acid and base
= heat taken up by calorimeter + solution
= M1 ´ S1DT + M2 + S2DT
= 12 ´ 1 ´ 4.4 + 600 ´ 1 ´ 4.4 = 2692.8 cals
\ Heat of nuetralisation = 5 ´ 2692.8 cals
= – 13.464 kcals
Illustration 3: 0.16 g of methane was subjected to combustion at 27°in a bomb calorimeter. The temperature of calorimeter system (including water) was found to rise by 0.5°C. Calculate the heat of combustion of methane at (i) constant volume (ii) constant pressure. The thermal capacity of the calorimeter system is
Solution: Heat evolved at constant volume = 0.16 ´ 17.7 ´ 0.5 kJ
Heat of combustion of methane (heat evolved per mole)
= = 885 kJ / mol
DE = – 885 kJ / mol
CH4(g) + 2O2(g) ¾® CO2(g) + 2H2O (l)
Dn = 1–3 = – 2
DH = DE + DnRT = – 885 + (–2) ´ 8.314 ´ 10–3 ´ 298
DH = – 889.95 kJ / mole
Illustration 4: 3.67 litres of ethylene and methane gaseous mixture on complete combustion at 25°C produces 6.11 litre of CO2. Find out the amount of heat evolved on burning one litre of gaseous mixture. The heats of combustion of C2H4 and CH4 are – 1423 and – 891 kJ mole–1 at 25°C.
Solution: Let volume of C2H4 in the mixture = x litre
Volume of CH4 in the mixture = ( 3.67 – x) Lit
C2H4 + 3O2 ¾® 2CO2 + 2H2O
CH4 + 2O2 ¾® CO2 + 2H2O
From above two equations, it is clear that x litre C2H4 will from 2x litre CO2 and (3.67–x) litre CH4 will from ( 3.67 –x) litre CO2
Hence, 2x + 3.67 – x = 6.11
x + 3.67 = 6.11
x = 2.44
\ Volume of CH4 percent in the mixture = 3.67 – 2.44 = 1.23 litre
Now, volume of C2H4 in one litre mixture = = 0.6648 litre
Volume of CH4 in one litre mixture = = 0.3352 litre
Above volumes are there at 25°C
Number of moles of methane PV = nRT
1 ´ 0.3352 = n ´ 0.0821 ´ 298 = n = 0.0137 mole
Heat evolved by combustion of methane
= Number of methane moles ´ Heat of combustion
= 0.0137 ´ 891 = 12.21 kJ
Number of moles of ethylene
PV = nRT
1 ´ 0.6648 = n ´ 0.0821 ´ 298
n = 0.02717 mole
Heat evolved by combustion of ethylene
= 0.02717 ´ 1423 = 38.68 kJ
Total heat evolved = 38.68 + 12.21
= 50.89 = 50.9 kJ.
Illustration 5: Calculate the proton affinity of NH3(g) from the following data (in kJ / mole)
DH°dissociation : H2(g) = 218
DH°formation : NH3(g) = – 46
Lattice energy of NH4Cl = – 683
Ionisation energy of H = 1310
Electron affinity of Cl = 348
Bond dissociation energy Cl2(g) = 124
Lattice energy NH4Cl(s) = – 314
Solution: We have to calculate DH for the following equation
NH3(g) + H+(g) ¾® (g) ————- (1)
Given; H2(g) ¾® 2H(g) : DH1 = 218 kJ / mole
N2(g) + H2(g) ¾® NH3(g) : DH2 = – 46 kJ / mol
NH4Cl (s) ¾® (g) + Cl–(g) : DH3 = + 683 kJ / mol H(g) ¾® H+(g) : DH4 = 1310 kJ / mol
Cl(g) ¾® Cl–(g) : DH5 = – 348 kJ / mol
Cl2(g) ¾® 2Cl (g) : DH6 = 124 kJ / mol
N2(g) + 2H2(g) + Cl2(g) ¾® NH4(s) DH7 = – 314 kJ / mol
DH = (DH1) – DH2 + DH3 – DH4 – DH5 – (DH6) + DH7
= –´ 218 + 46 + 683 – 1310 + 348 – ´ 124 – 314
= – 718 kJ / mol
Illustration 6: During the discharge of a lead storage battery, the density of H2SO4 falls from 1.294 to 1.139 g ml-1. H2SO4 of density 1.294 g ml-1 is 39% H2SO4 by weight and that of density 1.139 g ml-1 is 20% H2SO4 by weight. The battery holds 3.5 litres of acid and volume remains practically constant during the discharge. Calculate the number of ampere hours for which the battery must have been used. The charging and discharging reactions are:
Pb + ® PbSO4 + 2e– (charging)
PbO2 + 4H+ + SO42- ® PbSO4 + 2H2O (discharging)
Solution: The overall reaction is:
Pb + PbO2 + 2H2SO4 ® 2PbSO4 + 2H2O
It should be noted that 1 mole of Pb gets oxidised to 1 mole of Pb2+ ( in PbSO4). Therefore, during the process, 2 moles of H2SO4 are consumed with the utilisation of 2 moles of electrons. In other words, the number of moles of H2SO4 is equal to the number of Faradays consumed.
Mass of sample in 3.5 L of acid before the discharge = 3.5´1.294´103
= 4529 g
Mass of H2SO4 = 4529 ´ = 1766.31 g
Moles of H2SO4 = = 18.02
Mass of sample after discharge = 1.139 ´ 103 ´ 3.5 = 3986.5 g
Mass of H2SO4 = ´ 3986.5 = 797.3 g
No. of moles of H2SO4 = = 8.135
Moles of H2SO4 consumed = 18.02 – 8.135 = 9.88
\ No of Faradays utilised = 9.88 F
\Charge = 9.88 ´ 96500 = 953833.57 C
1 ampere hour = charge passed by one ampere in one hour
= 60 ´ 60 = 3600 C
\No of ampere hours = = 264.95
Illustration 7: Calculate the emf of the cell:
Pt H2 | CH3 COOH || NH4OH | H2Pt
1 atm. 0.1 M 0.01 M 1 atm.
Ka for CH3COOH = 1.8 x 10-5 and Kb for NH4OH = 1.8 x 10-5
Solution: The reactions are : at anode : ½ H2 ® + e–
at cathode : + e– ® ½ H2
——————————————————————— overall reaction ®
E = E0 – 0.059 log
From CH3COOH CH3COO– +
\= Ca = C
= 1.34 ´ 10–3 M
From NH4OH + OH–
[OH–] = C a =
\ = =
= = 2.359 ´ 10-11 M
\E = 0 – 0.0591 log = –0.4575 V
Illustration 7: The heat of solution in water was determined by measuring the amount of electrical work needed for compensate the cooling with would otherwise occur when the salt dissolves. After the NH4NO3 was added to the water, electrical energy was provided by passage of a current through a resistance coil until the temperature of solution reached the prior value. In a typical experiment, 4.4 gram NH4NO3 was added to 200 gm. water. A current of 0.75 ampere was provided through the heat coil and the voltage across the terminals was 6V. The current was applied for 5.2 minutes. Calculate DH for the solution of 1 mole NH4NO3 in enough water to give same concentration as was attained in above experiment.
Solution: Q = It = 0.75 ampere ´ 5.2 ´ 60S = 0.75 ´ 5.2 ´ 60 C
Energy = 0.75 ´ 5.2 ´ 60 C ´ 6V = 1.4 KJ
= mol = 0.055 mol
\ D = = 25 kJ mol–1
Illustration 9: How much time is required for complete decomposition of two moles of water using a current of 2 ampere?
Solution: 1 mol H2O º 2 eq. of H2O
\ 2 mol H2O º 4 eq. of H2O º 4F of electricity
º 4 ´ 96500 C
Time in second = = hr = 53.61 hr
Illustration 10: How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8 hr. at a current of 8.46 ampere? What is the area of a tray if the thickness of silver plating is 0.00254 cm. Density of silver is 10.5 gm cm–3.
Q = It = 8.46 amp ´ 8 ´ 3600 sec = 8 ´ 8.46 ´ 8 ´ 3600 C
Mass of silver = gram = 272.178 gm
Volume = = 25.92 cc
Surface area = = = 1.02 ´ 104 cm2
Illustration 11: 8.0575 ´ 10–2 kg of glauber’s salt (Na2SO4.10H2O) is dissolved in water to obtain 1 dm3 of a solution of density 1077.2 kgm–3. Calculate the molality, molarity and mole fraction of Na2SO4 in the solution.
Solution: Wt. of Glauber’s salt = w1 = 8.057 ´ 10–2 kg
= 80.575 gm
density of solution = d = 1077.2 kg m–3
Volume v = 1 L = 1000 ml
Molarity of Na2SO4 ×10H2O =
= 1L = 1 dm3 = 0.25 ml lit–1
= 0.25 mol dm–3 = 0.25 M
As we know,
Molality of Na2SO4 =
M1 = mol. wt of anhydrous salt Na2SO4 =
= 0.24 mol kg–1 solvant
» 0.25 M
For mole fraction of solute Na2SO4 (X1)
x1 + x2 = 1, where x2 = mole fraction of solvent
\ x2 = 1 – x1
(1 kg H2O = 1000 gm H2O = )
X1 = 0.0043
Illustration 12: The elements X and Y form compounds having molecular formula XY2 and XY4. When dissolved in 20 gm of benzene, 1 gm XY2 lowers the freezing point by 2.3°C, whereas 1 gm of XY4 lowers the freezing poing by 1.3°C. The molal depression constant for benzene is 5.1. Calculate the atomic masses of
X and Y.
Solution: Let the atomic mass of X = a
and the atomic mass of Y = b
For compared XY2
Mol. wt. = a + 2b
\mole of XY2 =
As we know that
DT¦ = K¦ ´ no. of moles of solute per 1000 gm solvant
or, DT¦ = k¦ ´
or, 2.3 = 5.1 ´
a + 2b = 110/869 …(1)
Similarly for compound XY4
m = mol. wt = a +4 b
D¦ = K¦ ´
1.3 = 5.1 ´ 196.153 …(ii)
Solving equation (i) and (ii), a = 25.60
b = 42.60
Illustration 13: Mixture of two liquids A and B is placed in cylinder containing piston. Piston is pulled out isothermally so that volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining, the mole fraction of A in vapour is 0.4. Given = 0.4 atm and = 1.2 atm at the experimental temperature. Calculate the total pressure at which the liquid has almost evaporated. (Assume ideal behaviour).
Solution: From Roult’s law
or, P = 0.4 × XA + 1.2 XB …(i)
i.e. Ya = 0.4, \ YB = 0.6
or, 0.4 = YA = …(ii)
YB = 0.6 = …(iii)
XA + XB = 1 …(iv)
From equation (ii), (iii) & (iv), wet get
XA = , XB =
Substituating the values of XA and XB in equation (i)
P = 0.4 ´ 2/3 + 1.2 ´ 1/3
= 0.268 + 0.396
= 0.664 atm
Illustration 14: A current of dry air was bubbled through a bulb containing 26.66 g of an organic substance in 200 gms of water, then through a bulb at the same temperature containing pure water and finally through a tube containing fused calcium chloride. The loss in weight of water bulb is 0.0870 gms and gains in weight of CaCl2 tube is 2.036 gm. Calculate the molecular weight of the organic substance in the solution.
Solution: As we know
Here A ® solvant, B ® solute
DP µ loss in wt. of water bulb
P0 µ gain in weight of CaCl2
From equation (1) & (2), we get
then mol. wt , mB = 63.59
Illustration 15: A dilute solution contains m mol of solute A in 1 kg of a solvent with molal elevation constant Kb; The solute dimerises in solution as 2A A2. Show that equilibrium constant for this dimer formation is K = , where DTb is the elevation in boiling point for the given solution. (Assume molarity = molality).
Solution: Let the degree of dimer formation be a. The reaction occurs as under:-
Initially m 0
after dimer formation m- ma
where m = molality = molarity (as also given)
Due to dimer formation,, the resultant
Molality will be =
Vont Hoff’s factor, i =
As we know the elevation in boiling point,
Þ DTb = Kb ´ i ´ Cm,
Þ DTb = Kb ´ ´ m, here Cm = m Þ a =
The equilibrium constant for dimer formations
Keq = =
Keq = …(1)
Putting the value of a in equation (1)
or, Kequ =
Illustration 16: The unit cell of a metal of atomic mass 108 and density 10.5 gm/cm3 is a cube with edge-length of 409pm. Find the structure of the crystal lattice.
Solution: r = , here Mm = 108
NA = 6.023 ´ 1023
Putting these values and solving, we get a = 409 pm
= 4.09 ´ 10–8 cm
r = 10.5 gm /cm3
n = 4 = number of atoms per unit cell
So, fcc lattice
Illustration 17: A compound alloy of gold and copper crystalizes in a cube lattice in which the gold atoms occupy the lattice points at the corners of a cube and the copper atoms occupy the centres of each of the cube faces determine the formula of this compound.
Solution: One – eighth of each corner atom (Au) and one – half of each face – centred atom (Cu) are contained within the unit cell of the compound.
Thus, number of Au atoms per unit cell = = 1 and number of Cu atoms per unit cell = 3. The formula of the compound is AuCu3.
Illustration 18: 5.35 g of a salt Acl (of weak base AOH) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827. Find the ionic radius of A+ and Cl– if ACl forms CsCl type crystal having 2.2 g/cm3. Given Kd(AOH)
= 1.8 ´ 10–5
= 0.732 for this cell unit
Solution: ACl(s) ¾¾® A+(aq) + Cl–(aq)
A+(aq) + H2O AOH(aq) + H+(aq)
at equilibrium C(1–a)
For salts of weak base and strong acid a =
H+ = Ca =
antilog (–4.827) =
M = 53.5
For CsCl type structure
= = 0.866 a
= 0.866 ´ 3.43 Å = 2.97 Å
Q = 0.732
on solving = 1.255 Å
= 1.715 Å
Illustration 19: In fcc arrangement of A & B atoms, where A atoms are at corners of the unit cell, B atoms at the face-centers, one of the atoms are missing from the corner in each unit cell then find the percentage of void space in the unit cell.
Solution: There areA atoms and 3 B atoms per unit cell
Also, 2 rA + 2 rB =
or, a =
Volume of unit cell = a3 =
Fraction of volume occupied per unit volume of the unit cell is given by
PF = =
So, void space = 1 – packing fraction = 1 – 0.570
= 0.430/unit volume of unit cell
\percentage of void space = 43%
Illustration 20: Prove that the void space percentage in zinc blende structure is 25%.
Solution: Anions are in fcc positions and half of the tetrahedral holes are occupied by cations. There are 4 anions and eight tetrahedral holes per unit cell.
Here, face-diagonal =
or, a =
r– = radius of anion
r+ = radius of cation
a = edge length of the cell
Volume of unit cell = a3 = =
\Packing Fraction = =
Since, for tetrahedral holes
= 0.7496 / unit volume of unit cell
\void space = 1 – 0.7496
= 0.2504/unit volume of unit cell
\percentage of void space = 25.04%
Illustration 21: Out of the following pairs, specify which of them are chiral-enantiomers or which of them are achiral-identical mirror images?
Solution: a) achiral – identical mirror images
- b) chiral – enantiomers
- c) chiral – enantiomers
- d) no chiral carbon – same structure
- e) chiral – enantiomers
- f) chiral – enantiomers
Illustration 22: What do you think about the following compounds – they are having chirality
Solution: a) It is optically active due to presence of chiral carbon.
- b) It has two chiral carbon atoms but due to plane of symmetry it is optically in-active.
- c) It has two chiral carbon atoms ()
hence it is chiral compound.
- d) There are no chiral carbon, but the molecule is chiral (an allene)
- e) There are no chiral carbon; this has a plane of symmetry – it is not a chiral molecule.
(f) Planar molecule – achiral
Illustration 23: Identify the end products in the following
This intermediate can
- i) Change to more stable free radical (3°) by 1,2 methyl shift to give other product.
which is not possible.
- ii) combine with another molecule to form higher alkane
Can you think about one other hydrocarbon?
Illustration 24: Explain mechanism of 1 : 4 addition of HBr on 1,3-butadiene.
Due to Mesomeric effect, there is polarity on C1 and C4 resulting in the 1,4 addition at these sites. This type of behaviour is generally obtained in conjugated systems.
Illustration 25: What are the products when HCl adds to 2,4- hexadiene?
First we decide which is more stable. Naturally (I) which is 2° allylic and in which p-electrons can be further delocalised. Hence addition of Cl– takes place on (I) in following way:
Illustration 26: Convert
- ii) PhH ¾¾®o-bromotoluene
Solution: i) One chlorine atom has been replaced by a carboxyl group. Hence, controlled conditions have to be used, and if we work backwards, two immediate possibilities are: (a) CuCN/H2O; (b) Grignard and carbon dioxide.
Illustration 27: Complete the following equations
Solution: i) Since the final product is an aldehyde, one precursor could be the 1,1-dichloride; these compounds are readily hydrolysed to carbonyl compounds.
There is also an alternative route.
- ii) Ph2C=CPh2
The use of two molecules of a 1,1-dichloride and copper suggests the formation of an alkene. This is an extension of the Wurtz reaction.
The use of the reagents given clearly indicates that a Grignard reaction is involved. However, aryl chlorides do not form Grignard reagents in ether; THF is necessary as solvent. On the other hand, aryl bromides (and iodides) readily react in ether solution.
The use of aqueous ammonium chlorides is a safety precaution to prevent the possibility of dehydration of the alcohol (which might occur if acid is used),
The use of iron indicates nuclear substitution. Me is o/p – orienting, and since there is always the possibility of a steric effect at the o-position, as we can suggest is that the 4-Br product will perdominate. Experimental work would have to be done to find out what are the actual results. From the literature it appears that the 4-Br compound is formed exclusively.
Illustration 28: Complete and comment:
Solution: The use of a strong base and a dihalogen derivative suggests the possibility of the formation of a methylene intermediate (compare with CH2Cl2 ¾® CHCl). This is supported by the formation of a three – membered ring compound by reaction with unsaturated compound . Hence, a possible
The product is triphenylcyclopropenium bromide.
Illustration 29: Write the mechanism of the following reaction
Tertiary butyl carbocation is Electrophile, which further attacks benzene ring. Many other reagents that can generate carbocations can be used for Friedel Craft’s reactions. Some of the reagents are
- Alkenes in presence of acids like BF3 or Mineral acid (HF).
- Alcohols in presence of acids like BF3 or Mineral acid (HF)
Identify A and also the Electrophile involved in the reaction.
(A) is iso propyl benzene or cumene and electrophile involved is CH3 – CHÅ – CH3 (Isopropyl Carbocation). Rearrangements are also possible in Friedel Crafts, reactions when benzene is treated with n-butyl chloride and Lewis acid the product obtained is isobutyl benzene.
- Naphthalene undergoes oxidation according to the reaction
C10H8(g) + 12O2(g) ® 10CO2(g) + 4H2O(l) and its heat of combustion is 1232.5 kCal. If the heats of formation of CO2 and H2O are 97 and 68.4 kcals respectively, calculate the heat of formation of naphthalene.
- Calculate the heat of formation of ethyl acetate from ethyl alcohol and acetic acid. Given that the heat of combustion of ethyl alcohol is 34000 cals, of acetic acid 21000 cals and of ethyl acetate 55400 cals
- Calculate the heat of formation of Ag2O from following data:
- i) Pb + 2AgNO3(aq.) ® Pb(NO3)2 (aq) + 2Ag + 509 cals
- ii) PbO + 2HNO3 (aq.) ® Pb(NO3)2 (aq) + H2O + 178 cals
iii) Pb + O2 ® PbO + 503 cals
- iv) Ag2O + 2HNO3 (aq) ® H2O + AgNO3 (aq) + 104 cals
- Calculate the heat of formation of As2O3 from data given below:
- i) As2O3 + 3H2O aq ® 2H3AsO3 (aq) – 7550 cals
- ii) As + 3Cl ® AsCl3 + 71.390 cals
iii) AsCl3 + 3H2O + aq ® H3AsO3 (aq) + 3HCl (aq) + 17580 cals
- iv) H + Cl ® HCl + 22000 cals
- v) HCl + aq ® HCl (aq) + 17315 cals
- vi) H2 + O ® H2O + 68360 cals
- Calculate resonance energy of CH3COOH from the following data if the observed heat of formation of CH3COOH is – 439.7 kJ.
Bond Energy (kJ) Heat of atomisation (kJ)
C – H = 413 C = 716.7
C – C = 348 H = 218.0
C = O = 732 O = 249.1
C – O = 351
O – H = 463
- The heat of formation of ethylene is 12.5 kcal. Calculate C = C bond energy in ethylene from following data.
Heat of atomisation of C = 170.9 kcal/mole
Heat of atomisation of H = 52.1 kcal/mole
Bond energy of C – H bond = 99.3 kcal/mole
- Ethylene undergoes combustion according to the thermochemical equation.
C2H4(g)+ 3O2(g) ® 2CO2(g) + 2H2O(l) DH° = –337 kcal
Assuming 70% efficiency how much water at 20°C can be converted into steam at 100°C by combustion of 103 litre of C2H4 gas at NTP. Heat of vapourisation of water at 20°C and 100°C are 1.00 kcal/kg and 540 kcal/kg respectively.
- The heat of formaiton of ethane is –20.3 kcal. Calculate the bond energy of C – C bond in ethane if the heats of atomisation of carbon and hydrogen are respectively 170.9 and 52.1 kcal per mole and bond energy of C – H bond is 99.0 kcal.
- A 1.00 L sample of a mixture of methane gas and oxygen measured at 25°C and 740 torr, was allowed to react at constant pressure in a calorimeter which, together with its contents, had a heat capacity of 1260 cal/K. The complete combustion of the methane to carbon dioxide and water caused a temperature rise in the calorimeter of 0.667 K. What was the mole percent of CH4 in the original mixture?
DH combustion for CH4(g) = –210.8 kcal/mol
- Using the data (all values are in kJ/mole at 25°C) given below:
(ethane) = -1559.8
(ethene) = –1410.9
(acetylene) = –1299.7
(acetaldehyde) = –1192.3
of CO2(g) = –393.5
of H2O (liq) = –285.8
for C(graphite) ® C(g) = 716.68
Bond energy of H – H = 435.94
Bond energy of O=O = 498.34
Calculate the following bond energies
(i) C – C (ii) C – H (iii) C = O
(iv) C = C (iv) C º C
- The commercial production of water gas utilizes the reaction under standard conditions: C + H2O(g) ¾® H2 + CO. The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to CO2. How many g of carbon must be burnt to CO2 to provide enough heat for the water gas conversion of 100g carbon. Neglect all heat losses to environment. of CO, H2O(g) and CO2 are –110.53, –241.81 and –393.51 kJ/mol respectively.
- In order to get maximum calorific output a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of fuel. A burner which has been adjusted for methane as fuel (with x litre/hour of CH4 and 6 x litre/hour of O2) is to be readjusted for butane C4H10. In order to get same calorific output, what should be the rate of supply of butane and oxygen. Assume that losses due to incomplete combustion etc., are the same for both fuels and that gases behave ideally. Heats of combustion:CH4 = 809 kJ/mol; C4H10 = 2878 kJ/mol.
- A solution of 5g of haemoglobin (M. Wt. = 64000) in 100 c.c of solution shows a temperature rise of 0.031°C for complete oxygenation. Each mole of haemoglobin binds four moles of oxygen. If the heat capacity of the solution is 4.18 JK–1 cm–3, calculate DH per g mole of oxygen bound.
- The (CaBr2(s)) = –675 kJ mol–1. The first and second ionization energies of Ca are 590 and 1145 kJ / mol. The enthalpy of formation of Ca is 178 kJ mol–1. The bond enthalpy of Br2 is 193 kJ mole–1 and enthalpy of vapourisation of Br2 is 31 kJ mol–1. The electron affinity of Br(g) is 325 kJ mole–1. Calculate the lattice energy of CaBr2(s).
- When 12.0 g of carbon reacted with oxygen to form CO and CO2 at 25°C and constant pressure, 75.0 kcal of heat was liberated and no carbon remained. Calculate the mass of oxygen needed for it and moles of CO and CO2 formed. Given DHf CO2 = –94.05 and DHf CO = –26.41 kcal mole–1
- Ammonium persulphate is prepared by the anodic oxidation of ammonium hydrogen sulphate to the reaction:
NH4HSO4 ® NH4SO4 + H+
2NH4SO4 ® (NH4)2 S2O8 + 2e– …. anode
2H+ 2e– ® H2 ….. cathode
H2O2 can be prepared by the hydrolytic reaction
(NH4)2 S2O8 + 2H2O ® 2NH4 HSO4 + H2O2
Assuming 60% current efficiency in the production of persulphate, calculate the amount of current necessary to produce 170 gms of H2O2. Yield in hydrolytic reaction is 80%.
- A current of 0.1 ampere is passed for 16 minutes through 100 ml of a 0.1 M NaCl solution, the cathode reaction taking place is 2e–+2H2O ® H2(g) + 2OH–
At the anode, Cl2 is liberated. Using this information, calculate the pH of the final solution after electrolysis.
- If E01 denotes the standard electrode potential of the electrode Fe/Fe2+ and E20 denotes the standard electrode potential for the electrode Fe/Fe2+/Fe3+, show that E03, the standard electrode potential of the electrode Fe/Fe3+ is given by the equation,
- Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained 0.1 M MnO–4 and 0.8M H+ and which has been treated with 90% of the Fe2+ necessary to reduce all the MnO–4 to Mn2+.
MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O E0 = 1.51 V
- An excess of liquid mercury was added to an acidified solution of 1´ 10-3 M Fe3+. It was found that only 4.6% of the iron remained as Fe3+ at equilibrium at 25°C. Calculate E0.
(Hg22+/Hg), assuming that the only reaction occurred was 2Hg + 2Fe3+ ®Hg22+
+ 2Fe2+. E0 (Fe3+/Fe2+) = + 0.771V.
- A cell consists of a half-cell containing a silver electrode immersed in saturated AgCl solution and another half-cell containing a silver electrode immersed in saturated Ag2CrO4 solution. If E0 (Ag+ | Ag) = 0.799 V, Ksp ( AgCl) = 1.7 ´ 10-10 and Ksp (Ag2CrO4)
= 1.9 ´ 10-12, determine the cell potential at 25°C.
- Kd for complete dissociation of Ag(NH3)+2 into Ag+ and NH3 is 6 ´ 10-8. Calculate E0 for the following half reaction : Ag(NH3)2+ + e– ® Ag+2NH3
Ag++e– ® Ag, E0 = 0.799 V.
- The overall formation constant for the reaction of 6 mol of CN– with cobalt (II) is 1 ´ 1019. The standard reduction potential for the reaction Co(CN)3-6 + e– ® Co (CN)+6 is 0.83 V. Calculate the formation constant of Co (CN)6 3-
Co3+ + e– ® Co2+; E0 = 1.82 V.
- For the cell Mg(s) | Mg2+(aq) || Ag+(aq) | Ag(s), Calculate the equilibrium constant at 25°C and the maximum work that can be obtained during operation of cell.
Given = 2.37 V and = + 0.80 V, R= 8. 314 J
- Two electrochemical cells are assembled in which the following reactions occur:
V2+ + VO2+ + 2H+ ® 2V3+ + H2O Ecell 0 = 0.616 V
V3+ + Ag+ + H2O ® VO2+ + 2H+ + Ag (s) E0 cell = 0.439 V
Calculate E0 for half reaction V3+ + e– ® V2+
Given = 0.799 V.
- For the galvanic cell Ag | AgCl(s) , KCl || KBr, AgBr(s) | Ag
0.2 m 0.001 m
Calculate the e.m.f. generated and assign correct polarity to each electrode for a spontaneous process after taking an account of cell reaction at 25°C. Given, = 2.8 ´ 10-10; = 3.3 ´ 10-13
- The standard reduction potential at 25°C for the reaction 2H2O + 2e– ® H2 + 2 0H– is
– 0.8277 V. Calculate the equilibrium constant for the reaction 2H2O ® H3O+ + OH–
- A lead storage cell is discharged which causes the H2SO4 electrolyte to change from a concentration of 34.6% by weight (density 1.261 g ml-1 at 25°C ) to one of 27% by weight. The original volume of electrolyte is one litre. How many Faradays have left the anode of battery ? Note, the water produced by the cell reaction as H2SO4 is used up. Over all reaction is:
Pb(s) + PbO2 + 2H2SO4 (l) ® 2PbSO4 (s) + 2H2O
- Zinc granules are added in excess to 500 ml of 1M Ni(NO3)2 solution at 25°C until the equilibrium is reached. If and are -0.75V and -0.24V respectively, find out the [Ni2+] at equilibrium.
- Calculate the minimum weight of NaOH required to be added in R.H.S. to consume all the H+ present in R.H.S. of cell of e.m.f. +710V at 25°C before its use. Also report the e.m.f. of cell after addition of NaOH.
= +0.760 V
- A 1.2% solution (wt./volume) of NaCl is isotonic with 7.2% solution (wt./volume) of glucose. Calculate degree of ionisation and Vant Hoff factor of NaCl.
- A solution containing 30g of a non-volatile solute in exactly 90g water has a vapour pressure of 21.85mm of Hg at 25°C. Further 18g of water is then added to solution, the new vapour pressure becomes 22.15 mm of Hg at 25°C. Calculate,
- a) Molecular weight of solute
- b) Vapour pressure of water at 25°C
- Liquids X and Y from an ideal solutions. The vapour pressures of pure X and Y at 100°C are 300 and 100 mm of Hg respectively. Suppose that the vapour, above a solution composed of 1 mole of X and 1 mole of Y at 100°C is collected and condensed. This condensate is then heated to 100°C and vapour is again condensed to form a liquid A. What is the mole-fraction of X in A?
- A solution containing compound X in water and a solution containing urea in water were put in a closed system. By doing this some water vapours were removed from one solution and got condensed in other. It is found where both the solutions were at equilibrium vapour pressure that one solution contains 2% of compound X and the other 5% urea by weight. Calculate the molecular weight.
- Calculate amount of that will separate out on cooling a solution containing 50g of ethylene glycol in 200g of water to –9.3°C (Kf for water = 1.86 kmol–1/kg)
- A certain liquid mixture of the liquids A and B behaving ideally shows a vapour pressure of 70mm Hg at 25°C for a certain mole fraction X of A. For the same mole fraction X for B in the mixture, the vapour pressure of the mixture is 90 torr at 25°C if the difference between the vapour pressure of the pure liquids (i.e. = 40 torr, calculate X, .
- A saturated solution of a sparingly soluble salt MCl2 has a vapour pressure of 31.78 mm of Hg at 30°C, while pure water exerts a pressure of 31.82 mm of Hg at the same temperature. Calculate the solubility product of the compound at this temperature.
- 1.25g benzoic acid (molar mass = 122) when dissolved in 100 cm3 of benzene produces osmotic pressure of 1.73 atm at 300 K. Benzoic acid is known to form dimer in benzene. Calculate (i) percentage of benzoic acid in the associated state (ii) the equilibrium constant of the dimerization reaction.
- Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) from a series of ideal solution over the whole range of composition. At 85°C, the vapour pressures of these pure liquids are 173 mm Hg and 127 mm Hg respectively.
- a) If 10g of ethylene dibromide is dissolved in 80g of 1, 2-dibromopropane. Calculate the partial pressures of each components and the total pressure of the solution at 85°C.
- b) Calculate the composition of the vapour in equilibrium with the above solution and express as mole fraction of ethylenedibromide.
- c) What would be the mole fraction of 1,2-dirbomopropane in solution at 85°C equilibriated with 50:50 mole mixture in the vapour?
- Two liquids A and B from ideal solutions. At 300°K, the vapour pressure of the solution containing 1 mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one mole of B is added to this solution, the vapour pressure of this solution increases by 10mm of Hg. Determine the vapour pressure of A and B in their pure state.
- When 11.7 g of sodium chloride are dissolved in 200gm of water, the depression in freezing point is double than the depression caused by 342 gm of cane sugar in 1000gm of water. From this information, what do you infer about the nature of solute particles in sodium chloride solution?
- An aqueous solution of glucose (M = 180) containing 12 gm dissolved in 100 gm water was found to boil at 100.34°C. The boiling point of pure water is 100°C. Calculate the molar constant for water.
- A solution of 0.643 g of an organic compound in 50 ml of benzene (density 0.879 g/ml) lowered its freezing point from 5.51°C to 5.03°C. Calculate the molecular weight of solid. Kf for benzene is 5.12 K mol–1 kg.
- Two elements A and B form compounds having molecular formula AB2 and AB4. When dissolved in 20g of benzene, 1g of AB2 lowers the freezing point by 2.3°C whereas 1g of AB4 lowers it by 1.3°C. The Kf for C6H6 is 5.1° mol–1 kg. Calculate atomic weight of
A and B.
- A mixture which contains 0.550g of camphor and 0.090 of an organic solute melts at 161°C. The solute contains 93.75% of C and 6.25% H by weight. What is the molecular formula of compound. K¢f for camphor is 37.5° mole–1 kg. The melting point of camphor is 209°C.
- Potassium metal crystallizes in a face-centered arrangement of atoms where the edge of the unit cell is 0.574 nm. What is the shortest separation of any two nuclei?
- An unknown metal is found to have a specific gravity of 10.2 at 25°C. It is found to crystallize in a body-centered cubic lattice with a unit cell edge length of 3.147Å. Calculate the atomic weight.
- The unit cell of tungsten is a face centered cubic having a volume of 31.699 Å3. The atom at the centre of each face just touches the atoms at the corners. Calculate the radius and atomic volume of tungsten.
- Compute the percentage void per unit volume of unit cell of Corundum.
- A metal Y – 47.9 crystallizes in hexagonal close packed structure in the unit cell, a = b = 295.3 pm with its height, h = 472.9 pm. Calculate its density.
- A spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP layers. The normal spinel has one-eighth of the tetrahbedral holes occupied by one type of metal ion and one-half of the octahedral holes occupied by another type of metal ion. Such a spinel is formed by Zn++, Al+3, and O– – with Zn+ + in the tetrahedral holes. Give formula of the spinel.
- A certain compound, whose density is 4.56 gm/cm3, crystallizes in the tetragonal system with unit cell distances of a = b = 658 pm and c = 593 pm. If the unit lattice contains four molecules, calculate the molar mass of the compound.
- Cesium chloride, bromide and iodide form interpenetrating simple cubic crystals like the other alkali halides. The length of the side of the unit cell of CsCl is 412.1± 0.3. Determine
- a) density of CsCl
- b) Ion radius of Cs+, assuming that the ions touch along a diagonal through the unit cell and that the ion radius of Cl– is 181 pm.
- The effective radius of an iron atom is 124 pm. Iron occurs both in a bcc and fcc structure. Calculate the ratio of densities of bcc-form and fcc-form.
- The density of CaF2 is 3.18 gm/cm3 at 20°C. Calculate the dimensions of the unit cube of the substance.
General Organic Chemistry
- Toluene on treatment with methyl chloride and AlCl3 at 0°C gives O – xylene and p – xylene, where as at 80°C the main product obtained is m – xylene, explain for the above deviation with a suitable mechanism ?
- What shape will you expect for the following molecules
- i) Amide ion ii) (CH3)3N
- Give reasons for the following
- a) Lithium acetyl acetonate has very higher melting point than beryllium acetyl acetonate.
- b) n – butyl alcohol has much higher boiling point than diethyl ether but both of them have same solubility in water.
- c) Why does every organic compound containing oxygen dissolves in Con H2SO4 and from the resulting solution the compound can be recovered by dilution with water.
- Explain the following observations.
- i) CH3 – I CH3OH + I–
- ii) CF3 – I CF3H + IO–
- Identify the configuration of products and also specify whether the products are optically active or inactive.
- a) 1- chloropentane + Cl2 (300°C) ¾® C5H10Cl2
- b) (S) –3- chloro –1- butene + HCl ¾® 2,3 dichloro 2-methyl butane
- c) (R) -2- chloro –2,3 – dimethyl pentane + Cl2 (300°C) ¾® C7H14Cl2
- Account for the following observations
- a) In benzene there are six equivalent carbon – carbon bonds, on the other hand in naphthalene (C1 – C2 bond length is 1.365Å, C2 – C3 is 1.404 Å) there are two bond lengths
- b) Acetylene is more acidic than benzene
- c) Aromatic bromination catalysed by Lewis acid thallium acetate gives only para isomer
- What is relation between following pairs of structure indicate whether they are identical or enatiomers.
- Which of the following compound is stronger acid and why?
Why in (i) enol form is more stable whereas (ii) mainly exists in keto form.
- Why is N, N-dimethyl –o-toluidine is stronger bas than N, N-dimethyl aniline?
Electrophilic Aromatic Substitution
- Complete the series of reaction
O —HOOC—C6H4—CH2—C6H5 ? ?
- Identify the compounds from A to D
- Predict the products of the following reactions.
- 0.450 g of an aromatic compound (A) on ignition gave 0.905 g of CO2 and 0.185g H2O 0.35 g of (A) on boiling with Nitric acid and on addition of AgNO3 solution gives 0.574 g of AgCl. The vapour density of (A) is 87.5. (A) on hydrolysis with Ca(OH)2 yields (B) which on mild reduction gives an optically active compound (C) On heating (C) with I2 and NaOH, iodoform is produced along with (D). Identify structures
from A to D.
- Predict the product in the reactions
- An organic compound (A) gives positive liberman reaction and on reaction with CHCl3/KOH followed by hydrolysis gives (B) and (C). Compound (B) gives colour with Schiff’s reagent but not (C), which is steam volatile. (C) on treatment with LiAlH4 gives (D), C7H8O2 which on oxidation gives (E). Compound (E), reacts with (CH3CO)2O/CH3COOH to give a pain reliever (F). Give structures (A) to (F) with proper reasoning.
- A neutral compound (A) C8H9ON on treatment with sodium hypo bromite forms an acid soluble substance C7H9N. On addition of aqueous sodium nitrite a solution of (B) in dilute HCl at 0-5°, an ionic compound (C) C7H7N2Cl is obtained. (C) gives red dye with alkaline b-naphthol solution when treated with potassium cupro cyanide, (C) yields a neutral substance (D) C8H7N. On hydrolysis (D) gives (E) C8H9O2. (E) liberates CO2 from aqueous sodium bicarbonate. (E) on oxidation with potassium permanganate gives (F) C8H6O4 (F) on nitration yields two isomeric mono nitroderivatives (G) and (H) having molecular formula C8H5NO6 identify A to H.
- The aromatic compound (A) C8H7Cl2 Br gives a pale yellow precipitate insoluble in dilute nitric acid, when warmed for a few minutes with alcoholic AgNO3 solution. The compound (A) on reaction with aqueous KOH yields (B) having molecular formula C8H8Cl2O. The compound (B) on reaction with I2 and NaOH a yellow precipitate (C) and a compound (D). Acidification of (D) with conc. HCl yields an organic acid (E) having equivalent weight 191. Heating of (E) with soda-lime gives (F) which forms a single mono-nitro derivative (G) on nitration. Give structures of (A) to (G) with reasons.
- Write the structures of the compounds from the following data.
- A hydrocarbon A (m.f. C9H10) upon complete hydrogenation requires 4 moles of H2 per mole of it. It gives a white precipitate with Tollen’s reagent. Treatment with NBS followed by alc.KOH gives hydrocarbon B (m.f. C9H8) which upon drastic oxidation and subsequent decarboxylation yields benzene. The compound A upon ozonolysis gives dioxal, formic acid and OHC—CH2—CH(CHO)—CH2COOH. Identify A and B and give the reactions involved in their identification. What will be final product when B is successively treated with NaNH2, CH3I and Li – liq. NH3
- 11.1 kcals
- –400 cals
- 68 cals
- 154680 cals
- –110.3 kJ / mole
- 14.05 kcal / mole
- 16.9 kg
- 80.7 kcal
- 10 mole percent
- i) C–C = 97.81 kJ
- ii) C–H = 454.64 kJ
iii) C=O = 804.26 kJ
- iv) C=C = 434.3 kJ
- v) CºC = 733.48 kJ
- 33.36 gm
- 5.48 (x) litre O2
- -41.47 kJ
- -2162 kJ mol–1
- 27.49 gm
- 2010416.7 Coulombs.
- pH = 12
- 1.39 V
- -0.798 V
- 0.37 V
- Kf= 6 ´ 1063.
- Log Kc = 107. 457,
Wmax = 6. 118 ´ 102 kJ
- = -0.256 V
- Ecell = – 0.037 V. Thus to get Ecell positive, polarity of cells should be reversed ie.
Cell is Ag ½AgBr(s) KBr÷÷ KCl AgCl, KCl/Ag 0.001m 0.2m and E = +0.037 V
- Kw = 1.25 ´ 10–4
- 1.255 Faraday
- [Ni2+] = 5.15 ´ 10-18 m
- W= 1.264 g, Ecell = 0.3765 V
- a = 0.95; i = 1.95
- a = 67.83; b = 23.78 mm
- 38.71 g
- x = 0.25; = 100; = 60 torr
- 5 ´ 10–5
- (i) 63.8% (ii) k = 23.56
- a) C2H4Br2 = 20.414 mm; C3H6Br2 = 111.88 mm
Total P = 132.301 mm
- b) 0.154; (c) 0.58
- A = 400 mm Hg; B = 600 mm Hg
- i = 2; ionised
- a = 25.59; b = 42.64
- 0.406 nm
- 6.3697 Å3
- 4.46 gm / cm3
- 175.8 gm
- (a) 3.99 gm / cm3
(b) 176.13 pm
55 5.46 Å
General Organic Chemistry
- At 0°C rate controlled product is obtained while at 80°C equilibrium controlled product i.e. m – xylene is obtained.
Although methyl group activates strongly at ortho and para positions, but it also favours dealkylation at these positions, but as meta isomer is formed more slowly and tends to persist with undergoing dealkylation HCl provides HÅ for reversal of alkylation.
In the case of smaller group like metyl, mechanism is slightly different
- i) Flat with tetrahedral angle
- ii) Pyramidal
- iv) Pyramidal
- v) Trigonal
- In the first case I is more electronegative than C and hence the positive charge is an the C atom. So OH– attacks C and displaces I–.
In the second case due to the presence of three F atoms, the electron deficiency of C increases. Hence it gives a positive charge on I. Hence OH– attacks I and displaces CF3.
By viewing the above structures the C1 – C2 bond is having double bond character in two structures and single bond in the remaining one the C – 2 bond is single bond in two structures and double in only one \ C1 – C2 has more double bond character than C2 – C3 and hence the bond is short.
- b) s–character is more in Acetylene than in benzene
- c) An acid base complex is formed, which transfers halogen without its electrons because of its bulkiness reagent attacks least hindered position.
- a,b,d enantiomers, C identical
- Carbanion which results from loss of H+ from (ii) is less stable because rings lose coplanarity due to steric repulsion and therefore are not involved in stabilization of negative charge through resonance whereas this is not possible in (i) where all rings are tied to each other.
- In rotation about the single bond is possible which helps in minimizing repulsion between two – C = O dipoles whereas in this rotation is not possible and so in order to reduce two – C = O dipole repulsion molecule prefers enol form which is also stabilized by hydrogen bonding.
- In N, N-dimethyl-o-toluidine due to steric repulsion from –N(CH3) group is pushed upward and does not remain in the plane of ring so that electrons on nitrogen are not involved in resonance with the ring, increasing availability of electrons on nitrogen and hence basic nature is enhanced.
Electrophilic Aromatic Substitution
- A = Toluene
B = p-toluidine
C = p-toluene diazonium chloride
D = HBF4
- A = C25H20 B = Biphenyl
C = CHI3
- a) A = C6H5CHBrCH2CH3 B = C6H5CH = CH – CH3
C = C6H5CH – CH – CH3 D = C6H5COOH + CH3COOH
G = C6H5 – C º C – MgX H = C6H5 – C º C – CH2C6H5
- Complete hydrogenation of A requires 4 mole of H2 per mole of A meaning 4 degree of unsaturation. Giving white precipitate with Tollen’s reagent indicate the presence of terminal alkynic group. Rest two degrees of unsaturation must be due to two alkenic linkage as evident from the products of ozonolysis. The formation of benzene after decarboxylation of the oxidation product of B suggests that B is a derivative of benzene. B has two it atoms less than A indicating thereby that during treatment with NBS and subsequent treatment with base has created a double bond in the molecule by dehydrohalogenation. Keeping all these facts in mind A and B may be assigned the structures as given below.
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