Question

If $8.3ml$ of a sample of $H_{2}S{O}_4\left(36N\right)$ is diluted by $991.7ml$ of water, the approximate normality of the resulting solution is:

• A. $0.4$
• B. $0.2$
• C. $0.1$
• D. $0.3$

Answer 1

The correct option is D $0.3$

Given,

Normality of $H_{2}S{O}_4\left(N_1\right)=36N$

Volume of $H_{2}S{O}_4$ $V_1=8.3ml$

Volume of Water=$991.7ml$

$\therefore$ Total volume $V_2=991.7+8.3=1000ml$

Now, for calculating Normality of resulting solutions $N_2$

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

$=36\times 8.3=N_2\times 1000$

$\Rightarrow N_2=\frac{36\times 8.3}{1000}=0.298\approx 0.3$