10 ml of a solution of H2O2 labelled ’10 volume’ just decolorises 100 ml of potassium permanganate solution acidified with dilute H2SO4. Calculate the amount of potassium permanganate in the given solution.

Q: 10 ml of a solution of H2O2 labelled ’10 volume’ just decolorises 100 ml of potassium permanganate solution acidified with dilute H2SO4. Calculate the amount of potassium permanganate in the given solution.

(A) 0.1563 gm

(B) 0.563 gm

(C) 5.63 gm

(D) 0.256 gm

Sol:  2H2O2 → H­2O + O2

22400 ml of O2 evolved from 68 gm of H2O2

∴ 10 ml of O­2 is evolved from $\frac{680}{22400}$ gm of H2O2

Hence 1 ml  of H2O2 contain  $ \frac{0.0303}{34} mol $ = 0.00178 equivalent 10 ml of H2O2 will 0.0178 equivalents which will be  present in 100 ml of KMnO4 solution.

Amount of KMnO­4 in given sample $=\frac{158}{5} \times 0.00178 $  = 0.563 gm

Ans: (B)

The hourly energy requirement of an astronaut can be satisfied by the energy released when 34g of sucrose are burnt in his body. How many g of oxygen would be needed to be carried in space capsule to meet his requirement for one day?

Q: The hourly energy requirement of an astronaut can be satisfied by the energy released when 34g of sucrose are burnt in his body. How many g of oxygen would be needed to be carried in space capsule to meet his requirement for one day?

(A) 916.2 gm

(B) 91.62 gm

(C) 8.162 gm

(D) 9.162 gm

Sol: C12H22O11 + 12O2  → 12CO2 + 11H2O

1 mole of starch requires 12 mole of oxygen

∴ $\large \frac{34}{342}$ mole of starch requires $\large \frac{34}{342} \times 12 \times 32 \times 24 $

= 916.2 gm

Ans: (A)

Calculate the weight of ion which will be converted into its oxide by the action of 18 g of steam on it.

Q: Calculate the weight of ion which will be converted into its oxide by the action of 18 g of steam on it.

(A) 37.3 gm

(B) 3.73 gm

(C) 56 gm

(D) 5.6 gm

Sol:

2Fe + 3H2O → Fe2O3

3 × 18gm of steam will react with 2 × 56 gm of Iron

∴ 18 gm of steam will convert  $\large = \frac{18 \times 2 \times 56}{3 \times 18} $

= 37.3 gm of Fe

Ans: (A)

25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl3. Calcualte the ratio of moles of ICl and ICl3.

Q:  25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl3. Calcualte the ratio of moles of ICl and ICl3.

(A) 1:1

(B) 1:2

(C) 1:3

(D) 2:3

Sol: I2 + 2Cl2 → ICl + ICl3

25.4 gm of Iodine contains 0.1 moles of it

14.2 gm of chlorine contains 0.2 moles of it

∴ Moles of ICl & ICl3 produced will be 0.1 and 0.1. Hence Molar ratio 1: 1

Ans:  (A)

The volume in ml of 0.1 N HCl required to react completely with 1.0 gm mixture of Na2CO­3 and NaHCO­3 containing equimolar amounts of the two compounds is

Q: The volume in ml of 0.1 N HCl required to react completely with 1.0 gm mixture of Na2CO­3 and NaHCO­3 containing equimolar amounts of the two compounds is

(A) 157.9 ml

(B) 152.6 ml

(C) 200 ml

(D) 98.5 ml

Sol:  Let amount of Na2CO3 in 1 gm of mixture be a and since Na2CO3 and NaHCO3 are in equimolar amounts.

$\large \frac{a}{106} = \frac{1-a}{84} $

∴  a = 0.588 gm

Weight of NaHCO3 = (1 – 0.558) = 0.442 gm

Met of Na2CO3 + Meq. of NaHCO3 = Meq. of HCl

$\large \frac{0.588}{53} \times 1000 + \frac{0.492}{84}\times 100 = 0.1 \times V$

∴ V = 157.9 ml

Ans:  (A)

A mixture of KBr and NaBr weighing 0.560 gm was treated with aqueous Ag+ and all the bromide ion was recovered as 0.970 gm of pure AgBr. The fraction by weight of KBr in the sample is

Q: A mixture of KBr and NaBr weighing 0.560 gm was treated with aqueous Ag+ and all the bromide ion was recovered as 0.970 gm of pure AgBr. The fraction by weight of KBr in the sample is

(A) 0.25 gm

(B) 0.2378

(C) 0.36

(D) 0.285

Sol:  KBr + NaBr          +  Ag+  →AgBr

a gm    (0.56 – a)                    0.97 gm

Applying POAc for Br atom 1 × nKBr  + 1 × nNaBr = 1 × nAgBr

$\large \frac{a}{119} + \frac{0.560 a}{103} = \frac{0.97}{188} $

∴  a = 0.1332 gm

∴ Fraction of KBr in the sample = 0.1332/0.560 = 0.2378

Ans: (B)

25 ml of H2O2 solution were added to excess of acidified solution of KI and iodine so liberated required 20 ml of 0.1 N Na2S2O3 for titration. The normality H2O2 is ?

Q: 25 ml of H2O2 solution were added to excess of acidified solution of KI and iodine so liberated required 20 ml of 0.1 N Na2S2O3 for titration. The normality H2O2 is ?

(A) 0.02

(B) 0.04

(C) 0.08

(D) 0.03

Sol:

O2–1 + 2e → 2O–2                   2 + S2O32–  → 2I + S4O62–

2I  → I2 + 2e

Meq. of H2O2  = Meq. of I2 = Meq of Na2S2O3

N × 25 = 0.1 × 20

$N_{H_2 O_2} = 0.08 $

Ans: (C)