Question

22.7 ml of N/10 $N{a}_{2}C{O}_3$ solution neutralizes 10.2 ml of a dilute $H_{2}S{O}_4$. Then the volume of water that must be added to 400 ml of same $H_{2}S{O}_4$ to make it exactly N/10 is:

• A. 245 ml
• B. 484.6 ml
• C. 480 ml
• D. 490 ml

Answer 1

The correct option is D 490 ml

Let normality of $H_{2}S{O}_4=N$

$N_1{V}_1=N_2{V}_2$

$\Rightarrow 0.1\times 22.7=N\times 10.2$

$\Rightarrow N=0.2225$

Now the number of equivalent present in

$400ml$ solution= $N\times 400$

$=0.2225\times 400=89meq$

For making the solution of $0.1N$, Let $V$ $ml$ of solution is needed

$\Rightarrow 0.1\times V=89$

$V=890ml$

Hence the volume of water that must be added is $400ml$

$\therefore V=890-400=490ml$