## CHEMISTRY

**Time: Two Hours Maximum Marks : 100**

* *

**Note:**

- i) There are
**TEN**questions in this paper. Attempt**ALL** - ii) Answer each question starting on a new page. The corresponding question number must be written in the left margin. Answer all parts of a question at one place only.

iii) Use of Logarithmic tables is permitted.

- iv) Use of calculator is
**NOT PERMITTED**

*Useful Data:*

Gas Constant **R ** = 8.314 J mol^{–1}K^{–1}

= 0.0821 lit atm mol^{–1} K^{–1}

= 2 Cal mol^{–1}

Avogadro’s Number **N** = 6.023 ´ 10^{23}

Planck’s constant **h** = 6.625 ´ 10^{–34} J sec.

Velocity of light **c** = 3 ´ 10^{8} m sec^{–1}

1 electron volt **ev** = 1.6 ´ 10^{–19} J

**F** = 96500 C

Atomic Masses Ag = 108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39,

Cl = 35.5, N = 14, S = 32, Na = 23, H = 1, P = 31, I = 127, As = 75,

Fe = 56, Ag = 108

** Name** :

** Enrolment No. **:

- a) An electron from an atom X de-excites from a state in which its s-function has 6 radial nodes to some other state, where its associated de-Broglie’s wavelength is minimum. Moreover, 20 ml of the gaseous atom X to pass through an orifice takes 5.0 secs. The diffusion of same ml oxygen gas through the same orifice takes 28.28 secs. If the frequency of light emitted is utilised in removing an e
^{–}from a metal atom Y whose work function is 1.08 eV, what is the kinetic energy of the ejected electron from the metal

atom Y?**[7]** - b) How much electric field strength should be applied in between two electric plates so as to produce an electron beam of same wavelength as that of the wavelength of light emitted from the atom X due to de-excitation from the states mentioned above.
**[5]**

- Calculate the time (in minutes) required for the solution to be optically inactive (following first order kinetics), from the following data.

t/min | 0.0 | 7.2 | 36.8 | 46.0 | ¥ |

Rotation of polarized light/degree (at 25°C) | 24.1 | 21.4 | 12.4 | 10.0 | –10.7 |

If the activation energy of this reaction is 15K cals/mole, at what time the solution become optically inactive, when the optical rotation is carried out at 40°C. (assume r_{o} is independent of temperature) **[10]**

- 3. a) The value of K
_{p}for the reaction

2H_{2}O(g) + 2Cl_{2}(g) 4HCl(g) + O_{2}(g) is 0.035 atm at 400^{o}C, when the partial pressures are expressed in atmosphere. Calculate K_{c} for the reaction,

Cl_{2}(g) + H_{2}O(g) **[5]**

- b) The density of an equilibrium mixture of N
_{2}O_{4}and NO_{2}at 1 atm. and 348 K is 1.84 gdm^{-3}. Calculate the equilibrium constant of the reaction, N_{2}O_{4}(g) 2NO_{2}(g).**[8]**

- Explain the following.
- a) CH
_{2}F_{2}involves sp^{3}hybridised C atom. Still none of the angles is 109.5° but is different from this. Why? - b) The boiling point of CH
_{4}is less than that of SiH_{4}but reverse is true for CCl_{4}and SiCl_{4}Why? - c) The bond energy of C=C bond is greater than that of –N=N– but the bond energy of

–CºC– is less than that of Nº Why? - d) Naphthalene undergoes oxidation more readily than benzene but only to the stage where a substituted benzene is formed. Further oxidation requires more vigrous conditions. Why?
**[4****´4]**

- Identify A,B,C,D,E and F

**[2 ****´ 6]**

- Bring out the following conversion

** [5]**

- By referring to the equilibrium constants listed below, find the equilibrium for each of the following:
- a) Fe(OH)
_{3}(s) + 3H_{3}O^{+}Fe^{3+}+ 6H_{2}O - b) PbSO
_{4}(s) + PbCrO_{4}(s) +

K_{sp}[Fe(OH)_{3}] = 4 ´ 10^{–38} , K_{w} = 10^{–14}, K_{sp}(PbSO_{4}) = 2 ´ 10^{–8}, K_{sp}[PbCrO_{4}] = 2 ´ 10^{–14}. **[10]**

- Arrange the following in the order of the property mentioned in brackets and explain your answer.
- a) NH
_{3}, CF_{3 }NH_{2}, CH_{3}NH_{2}, (CH_{3})_{2}NH (basicity) - b) SiO
_{2}, SO_{3}, N_{2}O_{5}, CO_{2}(acidity) - c) Ag
_{2}O, V_{2}O_{5}, CO, N_{2}O_{5}(acidity) - d) CH
_{4}, NH_{3}, H_{2}O & HF (strength of conjugate base)**[2.5****´4]**

- a) The police often use a device called a breath analyser to test drivers suspected of being drunk. For this process a sample of driver’s breath or blood is drawn into the breath analyser where it is treated with an acidic solution of K
_{2}Cr_{2}O_{7}. The limit of blood alcohol content is 0.1% by mass. If a 60 gm sample of blood from a driver required 28.64 ml of 0.07654 M K_{2}Cr_{2}O_{7}for titration, should the police prosecute the individual for drunken driving?**[4]** - b) Octane is a component of gasoline. Complete combustion of octane leads to CO
_{2}and H_{2}Incomplete combustion leads to CO and H_{2}O which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 2.000 gallon of octane is burned in an engine The total mass of CO, CO_{2}and H_{2}O produced is 23.06 Kg. Calculate the fraction of octane converted to CO_{2}. The density of octane is 2.650 Kg/ gallon.**[6]** - A weak base 50 ml was titrated with 0.1 M HCl. The pH of the solution after addition of 12 ml and 25 ml were found to be 9.84 and 9.24 respectively. Calculate K
_{b}of the base and pH - a) at equivalence point
- b) on addition of 20 ml HCl more after equivalence point.
**[10]**

vvv

** **

** **

## CHEMISTRY

**SOLUTIONS**

** **

- a) From the problem, we can infer that n
_{2}= 7 (7s has 6 radial nodes (n-l-1) and n_{1}= 1 (wavelength of electron is minimum). From the rate of diffusion data, we can find out the molecular weight of the gaseous molecule.

on solving, M_{x} = 1.00

i.e., x = hydrogen

K.E. of ejected electron from the metal atom Y

K.E. = hn – hn_{o} = =

= [13.32 – 1.08] eV (hn_{o} = 1.08eV)

K.E. = 12.24 eV

= 12.24 ´ 1.6 ´ 10^{–19} = 1.95 ´ 10^{–18} Joules

- b) l =

=

9.33 ´ 10^{–8}‑ =

933Å =

V =

V = 1.723 ´ 10^{–16} V

- = = =

= 0.0112

log

= 0.0149

The time at which, the solution becomes optically inactive at 40°C can be calculated as follows,

r_{t} = 0

t =

t = 79.16 minutes

i.e., after 79.16 minutes , the solution becomes optically inactive

- a) K
_{P}= K_{C}(RT)^{Dn}

Dn = moles of product – moles of reactants = 5 – 4 = 1

R = 0.082 L atm/mol K, T = 400 + 273 = 673 K

\ 0.035 = K_{C} (0.082 ´ 673)

K_{C} = 6.342 ´ 10^{-4} mol l^{-1}

\ for the reverse reaction would be

\ = = 1576.8 (mol l^{-1})^{-1}

When a reaction is multiplied by any number n (integer or a fraction) the or becomes (K_{C})^{n} or (K_{P})^{n} of the original reaction.

\ K_{C} for O_{2}(g) + 2HCl(g) Cl_{2}(g) + H_{2}O(g)

is **39.7 (mol.l ^{-1})^{-½}**

- b) Let us assume that we start with C moles of N
_{2}O_{4}(g) initially.

N_{2}O_{4}(g) 2NO_{2}(g)

Initially C 0

At equilibrium C(1-a) 2Ca

where a is the degree of dissociation of N_{2}O_{4}(g)

Since,

Initial vapour density =

Since vapour density and actual density are related by the equation,

V.D. = = = 26.25

\ 1 + a =

\ a = 0.752

\ K_{p} =

K_{p}= 5.2 atm.

- a) In this compound two types of angles are there one is HCH and another is FCF. The FCF angle is smaller than 109.5° because of higher electronegativity of F and due to this s – character decreases. On the other hand in HCH bond the s – character increases hence angle also increases from 109.5°.
- b) The electronegativity of Si is less than that C and due to this in SiCl
_{4}the Cl atom acquires more than twice as much negative charge and give rise to higher electrostatic repulsion among SiCl_{4}than among CCl_{4}molecule hence boiling point of SiCl_{4}is lower than that of CCl_{4}. - c) Because of the lone pair repulsion the -N=N- is weaker than C=C bond but the NºN bond is stronger than CºC bond because the lone pairs on the two nitrogen atoms are on the opposite sides and too far apart to interact strongly with each other.
- d) Resonance energy of naphthalene is 61 Kcal / mol and for benzene it is 36 Kcal/mol so energy required to destroy the first ring is (61-36)=25 Kcal and if once one ring is destroyed 36 kcal of energy is required which is greater than that for the 1st ring hence more vigrous conditions are required.

5. |

6. |

- a) 2H
_{2}O H_{3}O^{+}+ OH^{–}

K_{w} = [H_{3}O^{+}][OH^{–}] —— (1)

Fe(OH)_{3}(s) Fe^{3+} + 3OH^{–}

K_{sp} = [Fe^{3+}][OH^{–}]^{3} ————— (2)

cubing equation (1) and dividing equation (2), we get,

= ——— (3)

The left hand side of equation (3) is the equilibrium constant of equilibrium (a), hence

K_{c} = = = 4 ´ 10^{4}

- b) PbSO
_{4}(s) + PbCrO_{4}(s) +

K_{c} = —— (4)

PbSO_{4}(s) Pb^{2+} +

\ K_{sp}[PbSO_{4}] = [Pb^{2+}] ——- (5)

PbCrO_{4}(s) Pb^{2+} +

\ K_{sp}[PbCrO_{4}] = [Pb^{2+}] —— (6)

Dividing equations (5) and (6), we get,

K_{c} = = 10^{6}

- a) CF
_{3}NH_{2}< NH_{3}< CH_{3}NH_{2}< (CH_{3})_{2}NH - b) SiO
_{2}< CO_{2}< N_{2}O_{5}< SO_{3} - c) The oxide with the highest positive oxidation state on the element other than O should be most acidic. Oxidation states of V in V
_{2}O_{5}and N in N_{2}O_{5}are both +5. But the electronegativity of N is higher, making N_{2}O_{5}the most acidic oxide. - d) The size of central atom decreases from C to F, at the same time volume available to electron on the central atom also decreases because in CH
_{3}^{–}ion only 1/4th of the space is available for electrons. Thus increasing basic character of these conjugate bases is

F^{–} < OH^{–} < NH_{2}^{–} < CH_{3}^{–‑}

- a) 3CH
_{3}CH_{2}OH + 2K_{2}Cr_{2}O_{7}+ 8H_{2}SO_{4}¾¾® 3CH_{3}COOH + 2Cr_{2}(SO_{4})_{3}+ 2K_{2}SO_{4}+ 11H_{2}O

Moles of K_{2}Cr_{2}O_{7} = 28.64 ´ 10^{–3} ´ 0.07654

Moles of CH_{3}CH_{2}OH= ´ 28.64 ´ 10^{–3} ´ 0.07654 = 3.288 moles

= 3.228 ´ 46 ´ 10^{–3} gm of C_{2}H_{5}OH = 0.1512 gm

% by mass = = 0.25%

Since 0.25 > 0.1 the police should prosecute

- b) Mass of 2 gallon of octane = 2.650 ´ 2 ´ 1000 gm

Moles of octane = in 2 gallon

+ O_{2} ¾¾® +

+ ¾¾® +

8 ´ 44x + 9 ´ 18x + 8y ´ 28 + 9y ´ 18 = 23.06 ´ 10^{3} gm

x + y = 46.48 Þ x = 40.22

% = = ´ 100 = 86.5%

- Let BOH is base with normality N

BOH + HCl ¾® BCl + H_{2}O

Initial Meq. 50 N 10 ´ 0.1 0

Meq. of after titration 50 N – 1 0 1

Medium is still basic as clear from pH. So acid acts as limiting reagent. Further mixture above forms a basic buffer solutions.

pOH = pK_{b} + log

4.16 = pK_{b} + log …(1)

Similarly after addition of 25 ml HCl

4.76 = pK_{b} + log …(2)

From (2) – (1)

0.6 =

3.98 = Þ N = 0.1

putting N = 0.1 in equation (1)

4.16 = pK_{b} + log

K_{b}= 1.73 ´ 10^{–5}

At equivalence point we calculate pH by salt hydrolysis

B^{+} + H_{2}O BOH + H^{+}

C(1 – a) Ca Ca

C = = 0.05 (Total volume is doubled)

pH = (pK_{w }– pK_{b} – log C) = 5.12 = (14 – 4.76 – log 0.05) = 5.27

After addition of HCl, the pH is calculated from HCl left unreacted only

Meq. of HCl left = 20 ´ 0.1 = 2

Normality of HCl = = 0.0166; pH = – log (0.0166) = 1.78

** **