Question

The normality of 0.98 \%0.98% (w/v) \mathrm{H}_{2} \mathrm{SO}_{4}H
2
​ SO
4
​ solution is

(1)
0.1 \mathrm{N}0.1N
(2)
0.4 \mathrm{N}0.4N
(3)
0.2 \mathrm{N}0.2N
(4)
1N​

Answer 1

Question:

The normality of 0.98%(w/v) H₂SO₄ solution is

(1)  0.1 N

(2)  0.4 N

(3)  0.2 N

(4)  1 N​

Solution:

$Normality = \frac{Gram \ equivalent \ of \ solute}{Volume \ of \ solution \ in \ litre} = \frac{Weight}{Equivalent \ Weight} * \frac{1000}{V \ in \ ml}$

whereas,

$Equivalent \ Weight = \frac{Molar \ Mass }{n}$

Molar Mass of H₂SO₄ = 98 and as H₂SO₄ is a dibasic acid which means its basicity will be 2 or we can say that 1 mole of H₂SO₄ will give 2 moles of H⁺

$Equivalent \ weight \ of \ H_{2}SO_{4} = \frac{98}{2} = 49$

And in question 0.98 of H₂SO₄ is present in 100 ml of solution,

So,

$Normality = \frac{Weight}{Equivalent \ Weight} * \frac{1000}{V \ in \ ml}$

$Normality = \frac{0.98}{49} * \frac{1000}{100}$

$Normality = \frac{98}{49 * 100} * \frac{1000}{100}$

$Normality = \frac{2}{100} * 10$

$Normality = 0.2 N$

The normality of 0.98% H₂SO₄ solution is 0.2 N.

(3)  0.2 N is the correct answer.