Q: 22.7 ml of N/10 Na_{2}CO_{3} solution neutralises 10.2 ml of a dilute H_{2}SO_{4}. Then the volume of water that must be added to 400 ml of same H_{2}SO_{4} to make it exactly N/10 is

(A) 245 ml

(B) 484.6 ml

(C) 480 ml

(D) 490.2 ml

Sol: $\large \frac{22.7}{1000} \frac{1}{10} = \frac{10.2 \times N}{1000} $

Normality of H_{2}SO_{4} $\large = \frac{22.7}{10 \times 10.2} = \frac{22.7}{102}$

Let V ml of water be added to 400 ml $\frac{22.7}{102}$ NH_{2}SO_{4} to make it $\frac{N}{10}$ then,

$\large \frac{V+400}{1000} \times \frac{1}{10} = \frac{400 \times 22.7}{1000 \times 102}$

V + 400 = 890.2 ml

V = 490.2 ml

Ans: (D)