Chemistry objective questions

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Stoichiometry (Unsolved)

How many grams of phosphoric acid would be needed to neutralise 100gm of magnesium hydroxide?

(Molecular weight of H₃PO₄ = 98 and Mg(OH)₂ = 58.3 gm)

(A) 66.7 (B) 252gm

10 L of hard water required 0.56 gm of lime (CaO) for removing hardness. Hence temporary hardness in ppm of CaCO₃is

(A) 100 (B) 200

20 ml of xM HCl neutralises completely 10 ml of 0.1 M NaHCO₃ solution and further 5 ml of 0.2 M Na₂CO₃ solution in the presence of methyl orange at end point, the value of x is

(A) 0.167M (B) 0.133M

When 10 ml of ethyl alcohol (d = 0.7893 gm/ml) is mixed with 20 ml of water (d = 0.9971 g/ml) at 25°C, the final solution has a density 0.9571 gm/ml. The percentage change in total volume on mixing is

(A) 3.1% (B) 2.4%

0.635 gm of a cupric salt was dissolved in water and excess of KI was added in the solution and the liberated iodine required 25ml of N/5 Na₂S₂O₃ solution. The weight percentage of copper in the salt is

(A) 25% (B) 40%

The chloride of a metal (M) contains 65.5% of chlorine 100 ml of the chloride of the metal at STP weight 0.72gm. The molecular formula of the metal chloride is

(A) MCl₃ (B) MCl

22.4 litres of H₂S and 22.4 litre of SO₂ both at STP are mixed together. The amount of sulphur precipitated as a result of chemical calculation is

(A) 16gm (B) 23gm

In the mixture of NaHCO₃ and Na₂CO₃ volume of a given HCl required is x ml with phenolphthalein indicator and further y ml required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHCO₃ is

(A) 2x (B) y

A sample of oleum is labelled 109%. The % of free SO₃ in the sample is

(A) 40% (B) 80%

7.65 ´ 10^–3 moles of NaNO₃ require 3.06 ´ 10^–2 moles of a reducing agent to get reduced to NH₃. What is the n-factor of the reducing agent?

(A) 2 (B) 3

How many grams of sodium bicarbonate are require to neutralise 10ml of 0.902 M vinegar?

(A) 8.4 gm (B) 1.5gm

1 gm mole of oxalic acid is treated with conc. H₂SO₄. The resultant gaseous mixture is passed through a solution of KOH. The mass of KOH consumed will be

(A) 28gm (B) 56gm

NH₃ + OCl^– ¾® N₂H₄ + Cl^–

On balancing the above equation in basic solution using integral coefficients, which of the following whole number will be the coefficient of N₂H₄?

(A) 1 (B) 2

A certain compound has the molecular formula X₄O₆. If 10 gm of X₄O₆ has 5.72g X, atomic mass of X is

(A) 32 amu (B) 37 amu

In the following reaction

28NO₃^– + 3As₂S₃ + 4H₂O ¾® 6AsO₄^3– + 28NO + 9SO₄^2– + 8H⁺

equivalent weight of As₂S₃ (with molecular weight M) is

(A) (B)

Answers

C 2. B 3. C
A 5. C 6. A
C 8. D 9. A
A 11. C 12. D
A 14. A 15. D

Solved Objective

Problem 1: One mole of a mixture of CO and CO₂ requires exactly 20 gms of NaOH to convert all the CO₂ into Na₂CO₃. How many more gms of NaOH would it require for conversion into Na₂CO₃ if the mixture (one mole) is completely oxidised to CO₂.

(A) 60 gm (B) 80 gm

Solution: Moles of NaOH =

Moles of CO₂ = [Q ‘n’ factor for CO₂ = 2]

Moles of CO =

Moles of CO₂ produced = from CO

Moles of NaOH extra =

Mass of NaOH extra = 60

Problem 2: One litre of 0.1 M CuSO₄ solution is electrolysed till the whole of copper is deposited at cathode. During the electrolysis a gas is released at anode. The volume of the gas is

(A) 112ml (B) 254 ml

Solution: When copper is deposited at cathode, oxygen gas is released at anode

Equivalents of copper = 0.1 ´ 2 = 0.2

Equivalents of oxygen = 0.2

Volume of oxygen at NTP = 0.2 ´ 5600 = 1120 ml

Problem 3: In a reaction

FeS₂ + MnO₄ + H⁺ ¾® Fe³⁺ + SO₂ + Mn²⁺ + H₂O

The equivalent mass of FeS₂ would be equal to

(A) Molar mass (B)

Solution: Fe²⁺ ¾® Fe³⁺ + e^–

S₂^–2 ¾® 2S⁺⁴ + 10e^–

–––––––––––––––––

FeS₂ ¾® 2S⁺⁴ + Fe³⁺ + 11e^–

\ Equivalent mass of FeS₂ =

Problem 4: Equal volumes of 0.2M HCl and 0.4M KOH are mixed. The concentration of the principal ions in the resulting solution are

(A) [K⁺] = 0.4M, [Cl^–] = 0.2M, [H⁺] = 0.2M

(B) [K⁺] = 0.2M, [Cl^–] = 0.1M, [OH^–] = 0.1M

(D) [K⁺] = 0.2M, [Cl^–] = 0.1M, [OH^–] = 0.2M

Solution: Resulting solution in alkaline since M(KOH) > M(HCl), hence the molarity of KOH after reaction

= = 0.1M

KOH + HCl KCl + H₂O

t = 0 0.4 0.2

t = t 0.2 0 0.2

0.1 0.1 due to dilution

\ [K⁺] = 0.1 + 0.1 = 0.2M

[Cl^–] = 0.1M

[OH^–] = 0.1M

Problem 5: To prepare a solution that is 0.5M KCl starting with 100ml of 0.4M HCl

(A) Add 0.75gm KCl (B) Add 20 ml of water

Solution: a) 100ml of 0.4M KCl = mol = 0.04 mol (initial)

= 0.04 ´ 74.5 (initially) = 2.98gm

= 3.73g (after) = 0.05 mol in 100 ml

= 0.5M

\ (A) is true

b) Addition of 20 ml water will decrease molarity that will be less than 0.4 M

\ (B) is false

c) 04 (initial as in (a)) + 0.10 (added)

= 0.14 mol KCl in 100 ml

= 1.4 M

\ (C) is false

d) 04 mol KCl in 90 ml solution (after 10 ml water evaporated) = 0.044 M

\ (D) is also false

Problem 6: For the reaction

+ ¾® CaHPO₄ + 2H₂O

Which are true statements

(A) Equivalent weight of H₃PO₄ is 49

(B) Resulting mixture is neutralised by 1 mol of KOH

(D) 1 mol of H₃PO₄ is completely neutralised by 1.5 mol of Ca(OH)₂.

Solution: a) By given reaction

H₃PO₄ º 2OH^–

\ Equivalent weight = = 49

\ (A) is true

b) Only one acidic H in CaHPO₄ hence neutralised by 1 mol of KOH

\ (B) is true

c) One acidic H in CaHPO₄ makes it acidic salt.

\ (C) is true

d) H₃PO₄ º 3H⁺ º 3OH^– º5 Ca(OH)₂

\ (D) is true

Problem 7: The brown ring complex compound is formulated as [Fe(H₂O)₅NO]SO₄. The oxidation number of iron is

(A) 1 (B) 2

Solution: [Fe(H₂O)₅NO]SO₄ [Fe(H₂O)₅NO]²⁺ + SO₄^2–

In this complex NO transfers one of its electron to Fe (which is initially +2)

This gives +1 charge on NO and +1 charge on Fe. Fe has thus three unpaired electrons as confirmed by its magnetic moment which is B.M.

\ (A)

Problem 8: An organic compound contains 4% sulphur minimum molecular weight is

(A) 200 (B) 400

Solution: 4 gm sulphurs is in 100g compound hence 32gm sulphur is in = 800 g compound.

\ (C)

Problem 9: If 20 ml of 0.2 M K₃[Fe(CN)₆] is reduced by some equivalents of N₂H₄, then calculate numbers of moles of N₂H₄ required for the above reaction the answer is

(A) 10^–5 (B) 10^–3

Solution: N₂H₄ + K₃[Fe(CN)₆ + KOH ¾® K₄[Fe(CN)₆] + N₂ + H₂O

\ Number of equivalents of

K₃[Fe(CN)₆] =

Number of moles of

N₂H₄= = 10^–3

\ (B)

Problem 10: 0.7gm of Na₂CO₃.XH₂O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralisation. The value of x is

(A) 7 (B) 3

Solution: Meq. of Na₂CO₃.xH₂O in 20 ml = 19.8 ´

\ Meq. Of Na₂CO₃.xH₂O in 100 ml = 19.8 ´ ´ 5

\ = 19.8 ´ ´ 5

or =

M = 141.41

\ 23 ´2 + 12 + 3 ´ 16 + 18x = 141.41

x = 2

\ (C)

Chemical Equilibrium

For the reaction A + B C + D, equilibrium concentrations of [C] = [D] = 0.5M if we start with 1 mole each of A and B. Percentage of A converted into C if we start with 2 mole of A and 1 mole of B is

(A) 25% (B) 40%

CH₃ – CO – CH_3(g) CH₃ – CH_3(g) + CO_(g)

Initial pressure of CH₃COCH₃ is 100mm when equilibrium is set up, mole fraction of CO_(g) is hence K_p is

(A) 100mm (B) 50mm

For the reaction

CaCO_3(s) CaO_(s) + CO_2(g)

Equilibrium constant K_p is 1.642 atm at 1000K, if 40 gm of CaCO₃ was put into a 10 L flask, percentage of calcium carbonate would remain unreacted at the equilibrium.

(A) 66% (B) 34%

If equilibrium constant of

CH₃COOH + H₂O CH₃COO^– + H₃O⁺

Is 1.8 ´ 10^–5, equilibrium constant for

CH₃COOH + OH^– ¾® CH₃COO^– + H₂O is

(A) 1.8 ´ 10^–9 (B) 1.8 ´ 10⁹

PCl₅ is 40% dissociated when pressure is 2 atmosphere it will be 80% dissociated when pressure is approximately

(A) 0.2 atm (B) 0.5 atm

For the reaction: 2HI_(g) H_2(g) + I_2(g), the degree of dissociated (a) of HI_(g) is related to equilibrium constant K_p by the expression

(A) (B)

For which of the following reactions, the degree of dissociation cannot be calculated from the vapour density data.
i) 2HI_(g) H_2(g) + I_2(g)
ii) 2NH_3(g) N_2(g) + 3H_2(g)

iii) 2NO_(g) N_2(g) + O_2(g)

iv) PCl_5(g) PCl_3(g) + Cl_2(g)

(A) (i) and (iii) (B) (ii) and (iv)

Van’t Hoff equation giving the effect of temperature on chemical equilibrium is represented as

(A) (B)

Pure ammonia is placed in a vessel at temperature where its degree of dissociation (a) is appreciable at equilibrium.

(A) K_p does not change with pressure (B) a does not change with pressure

Steam starts reacting with iron at high temperature to give hydrogen gas and Fe₃O₄. The correct expression for equilibrium constant is

(A) (B)

X nY, X decomposes to give Y (in one litre vessel) if degree of dissociation is a then K_C and its unit.

(A) ,mol^n–1 lit^n–1(B) ,molⁿ litⁿ

Solubility of a solute in a solvent (say H₂O) is dependent on temperature as given by

S = Ae^–^DH/RT where DH is heat of reaction

Solute + H₂O Solution, DH = ±X

For a given solution variation of log S with temperature is shown graphically. Hence solute is

(A) CuSO₄.5H₂O (B) NaCl

For the reaction A_(g) B_(g)+ C_(g)

(A) K_p = a³P (B) K_p = a²(K_p + P + 1)

The values of K_C for the following reactions are given as below

A B, K_C =1, B C, K_C = 3 and C D, K_C = 5

Evaluate the value of K_C for A D

(A) 15 (B) 5

For the reaction (1) and (2)

A B + C

D 2E

Given: : : 9 : 1

If the degree of dissociation of A and D be same then the total pressure at equilibrium (1) and (2) are in the ratio.

(A) 3:1 (B) 36:1

ANSWERS

D 2. B 3. B
B 5. A 6. D
A 8. C 9. A
C 11. D 12. D
C 14. A 15. B

Gaseous State

At constant pressure what would be the percentage decrease in the density of an ideal gas for a 10% increase in the temperature.

(A) 10% (B) 9.1%

A small hole is pricked in a container containing a mixture of CH₄ and O₂ and placed in vacuum. What should be the molar ratio of methane to oxygen in the container when the two gases are effusing out in the ratio of 2:1

(A) (B)

The Van der Wall’s equation is (V – nb) = nRT for n moles where ‘a’ and ‘b’ are Van der Wall’s constant which of the following statements are true about ‘a’ and ‘b’ when the temperature of the gas is too low

(A) Both remains same (B) ‘a’ remains same, b varies

The tube in the figure is shielded at both end and heated upto double the original temperature both side of Hg column gases are packed with increasing temperature the Hg column

(A) Shift towards ‘B’ (B) Shifts towards A

the molar specific heat at constant volume of mixture of gases A and B is 4.33 cal. A is monoatomic and B is diatomic then the ratio of moles of A and B is

(A) 1:1 (B) 2:1

Rate of diffusion of two gases are r_A and r_B, molecular weights are M_A and M_B then partial pressure P_Ais (if number of moles are n_A and n_B).

(A) (B)

There are two gases A and B having degree of freedom f_A and f_B, the difference in energy supplied for increment in temperature of one mole gas by 50°C is (in cal)

(A) 50 (f_A – f_B) (B) 100 (f_A– f_B)

When an ideal gas undergoes unrestricted expansion no cooling takes place because the molecules

(A) Exert no attractive forces on each other (B) Do work equal to loss of KE

If two gases of molecular weights M_A and M_B at temperatures T_A, T_B and T_AM_B= T_BM_A, then which property has the same magnitude for both the gases.

(A) Density (B) Pressure

At low pressure Vander Wall’s equation for 3 moles of a real gas will have its simplified form

(A) (B)

for two gases A and B with molecular weights M_A and M_B, its observed that a certain temperature T, the mean velocity of A is equal to the root mean square velocity of B. Thus the mean velocity of A can be made equal to the mean velocity of B if

(A) A is at temperature T, and B at T¢×T > T¢

(B) A is lowered to a temperature T₂ = T

(D) Both A and B are placed at lower temperature

The quantity represents the

(A) Number of molecules in the gas (B) Mass of the gas

The rms velocity of hydrogen is times the rms velocity of nitrogen. If the temperature of the gas

(A) T(H₂) = T(N₂) (B) T(H₂) > T(N₂)

1 lt of N₂ and l of O₂ at the same temperature and pressure were mixed together what is the relation between the masses of two gases in the mixture.

(A) (B)

Consider a mixture of SO₂ and O₂ kept at room temperature compared to the oxygen molecule, the SO₂ molecule will hit the wall with

(A) Smaller average speed (B) Greater average speed

ANSWERS

B 2. A 3. D
C 5. C 6. A
A 8. A 9. D
A 11. B 12. A
C 14. C 15. A

Gaseous State (Solved)

Problem 1: The Vander Wall’s constant b is equal to

(A) The molecular volume of 1 mol of the gas

(B) Two times the molecular volume of 1 mole of the gas

(D) Four times the molecular volume of 1 mole of the gas

Solution: ‘b’ is the volume occupied by 1mole of a gas molecules. Excluded volume for 2 gas molecules

= p (2r)³ = pr³

\ Excluded volume / molecule =

\ b = ´ N_A

\ b = 4 ´ volume of 1 mole of molecules

Problem 2: What will be percentage of ‘free volume’ available in 1 mole of H₂O_(l) at 760 mm and 373 K. Density of H₂O_(g) at 373 K is 0.958 gm/ml.

(A) 0.0613% (B) 99.9386%

Solution: Volume occupied by 1 mole of H₂O_(g) at 373 K =

= = 30.62L

Volume of 1 mole of H₂O_(l)

= = 18.789 ml

% of volume occupied by liquid water

= 0.0613%

% of free volume

= 100 – 0.0613 = 99.9386%

Problem 3: One mole of each monoatomic, diatomic and triatomic gases are mixed, C_p/C_V for the mixture is

(A) 1.40 (B) 1.428

Solution: C_v =

C_p =

= 1.428

Problem 4: The vapour pressure of water at 20°C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20°C and 40% relative humidity.

(A) 4.2 ´ 10^–4 mole (B) 4.2 ´ 10^–6 mole

Solution: Relative humidity (RH) =

\ Partial pressure of H₂O = RH ´ Vapour pressure of H₂O

= = 7 torr

= = 0.0092 atm

Now PV = nRT

= 0.000382 mole

= 3.82 ´ 10^–4 mole

Problem 5: The compressibility factor of a gas is less than unity at STP. Therefore

(A) V_m > 22.4 litre (B) V_m < 22.4 litre

Solution: Z =

Ar Z < 1 (given)

\ or PV < nRT

P = 1 atm

V = V_m

R = 0.082

T = 273

\ V_m < 0.0821 ´ 273

= V_m < 22.4 litre

Problem 6: The composition of the equilibrium mixture (Cl₂ 2Cl), attained at 1200°C, is determined by measuring the rate of effusion through a narrow aperture. At 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as Krypton under the same conditions. What will be the fraction of chlorine molecules dissociated into atoms (Atomic mass of Kr = 84)

(A) 13% (B) 13.7%

Solution:

or M = 62.425

for Cl₂ 2Cl

initially 1 0

at equilibrium (1 – a) 2a

i (Van’t Hoff factor) = 1 – a + 2a = 1 + a

or 1 + a =

on solving a = 0.137

or a = 13.7%

Problem 7: A spherical balloon of 21 cm diameter and 4.851 L volume is to be filled with hydrogen at NTP from a cylinder containing the gas at 20 atm and 27°C the cylinder can hold 2.82 litre of water at N.T.P. what will be the number of balloons that can be filled up?

(A) 5 (B) 10

Solution: Volume of one balloon = 4.821 litre

Volume of n balloons = 4.821 n litre

Total volume of hydrogen in the cylinder at NTP

V = = 51.324 litre

Actual volume of H₂ to be transferred to balloons

= 51.324 – 2.82 = 48.504 litre

(\ 2.82 L is retained in the cylinder)

\ NO. of balloon, n = » 10

Problem 8: The numerical value of for a gas at critical condition is … times of at normal condition

(A) 4 (B) 8/3

Solution: At critical temperature

\ P_CV_C= RT_C

\ for a gas at critical temperature is times of that gas at NTP

Problem 9: What will be the molecular diameter of Helium if Vander Waal’s constant, b = 24 ml mole^–1?

(A) 2.71Å (B) 2.71mm

Solution: b = 4V_m

(V_m is the volume occupied by one mole of gas)

or 24 = 4 ´ N₀ ´ pr³

or r =

= 1.355 ´ 10^–8 cm

\ d = 2r = 2 ´ 1.355 ´ 10^–8 cm

= 2.71Å

Problem 10: the density of phosphorus vapour at 310°C and 775 torr is 2.64g dm^–3. What is the molecular formula of phosphorus?

(A) P (B) P₂

Solution:

or M =

Put d = 2.64gm dm^–3

R= 0.0821 dm³ atm K^–1 mol^–1

P = atm

and T = 310 + 273 = 583K

We get M = 123.9g mol^–1

No. of P atoms in a molecule

= = 3.99» 4

Thus the molecular formula of phosphorus = P₄

Electrophilic Aromatic Substitution (UNSOLVED)

1.	A and B are
(A)		(B)
(C)		(D)

Benzene diazonium chloride most readily couples with

(A) Anisole (B) Naphthalene

3.
(A)			(B)
(C)			(D)

4.	So A is
(A)		(B)
(C)		(D)

(A) Et– (B) Me–

6.
(A)			(B)
(C)			(D)

What is the name of the reaction?

(A) Fries rearrangement (B) Claisen rearrangement

Which of the following can react with TsCl?

(A) Glycerol (B) n-propyl cyanide

C* is with the product

(A)	CO₂	(B)
(C)	Both	(D)	None

10.
(A)	(B)
(C)	(D)	No reaction

11.
(A)
(B)
(C)
(D)		Any of these

12.
(A)			(B)
(C)			(D)

13.	Product is
(A)			(B)
(C)		Both	(D)	None

14.
(A)		(B)
(C)	Both are correct	(D)	None is correct

End product of the following reaction is

(A)		(B)
(C)		(D)

Answers

C 2. C 3. B
C 5. C 6. A
B 8. A 9. A
B 11. B 12. A
C 14. A 15. B

Solved Problems

Problem 1:
(A)			(B)
(C)			(D)

Solution:

\ (B)

Problem 2:	A and B are
(A)
(B)
(C)
(D)		None is correct

Solution:

\ (A)

Problem 3: The products of ozonolysis of two xylenes I and Ii are

(A) Same (B) Different

Solution:

o-xylene gives a mixture of glyoxal, methyl glyoxal and dimethyl glyoxal due to different positions of double bonds in the ring.

\ (B)

Problem 4:	The product is
(A)
(B)
(C)
(D)		Oxidation is not possible

Solution: An alkyl side chain attached to the benzene ring can be oxidised to –COOH group only when it contains atleast one a-hydrogen atom. Since t-butyl group has no a-H atom, it can not be oxidised, instead of this ring is oxidised.

Problem 5: When benzene in excess reacts with CH₂Cl₂ in presence of anhydrous AlCl₃ the product formed is

(A)		(B)
(C)		(D)

Solution:

Problem 6: The acid undergoes decarboxylation readily is

(A)		(B)
(C)		(D)

Solution: b-keto acids undergo spontaneous decarboxylation on heating to form a ketone.

Problem 7: The conversion which can be brought about under Wolff Kishner reduction conditions is:

(A) Benzaldehyde into benzyl alcohol

(B) Benzophenone into diphenyl methane

(D) Cyclohexanal into cyclohexanone

Solution: Under Wolf Kishner reduction a >C = O group is converted to >CH₂

Problem 8: The product obtained by heating salicylaldehyde with ethanoic anhydride in presence of sodium ethanoate is

(A)		(B)
(C)		(D)

Solution:

Problem 9: The intermediate formed in the following reaction is

(A) Carbene (B) Nitrene

Solution: The reaction is an example of cine substitution or elimination addition

Problem 10: A new carbon – carbon bond is formed in:

Cannizaro reaction Freidal-Craft’s reaction

(I) (II)

Clemmensen’s reduction Riemer-Tiemann reaction

(III) (IV)

(A) (II) (B) (II) and (IV)

Solution: New carbon-carbon bond is formed is Friedal Craft reaction and Reimer-Teimann reaction

Friedal Craft’s reaction

Riemer Teimann Reaction

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