Question

if 8.3 ml of sample of H2SO4 (36 N) is diluted by 991.7 ml of water , the approximate normality of resulting soltionis

Answer 1

V1N1= V2N2
V2 = 8.3 + 991.7
putting the values

8.3 x 36 = 1000 x N2
N2= 8.3 x 36/ 1000
= 0.2988N

Answer 2

Molecular weight of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is 98 gm. Its valence is 2. Therefore, the equivalent weight of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is Molecular weight divided by valence $=\left(\frac{98}{2}\right) gm=49 gm$.  

The quantity of $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution of 36 N to be taken is $8.3 \mathrm{ml}=\frac{8.3}{1000}=0.0083$.

Therefore, the amount of gram equivalent of $\mathrm{H}_{2} \mathrm{SO}_{4}$ present in 0.0083 of 36 N solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is $\begin{array}{l}{\text {= normality } \times \text { gram equivalents } \times \text { volume of solution taken }\\=36 \times 49 \times 0.0083} \\ {=14.6412 \mathrm{gm}}\end{array}$

Normality of the resulting solution when 8.3ml of 36N $\mathrm{H}_{2} \mathrm{SO}_{4}$ is diluted in 991.7ml  

$\left(=\frac{991.7}{1000}=0.9917\right)$ of water is:

$\begin{array}{l}{=\text { quantity of } \mathrm{H}_{2} \mathrm{SO}_{4} \text { present in grams } / \text { (equivalent weight of } \mathrm{H}_{2} \mathrm{SO}_{4} \times \text { volume of }} \\ {\text { water in which it is diluted in } )}\end{array}$

$=\frac{14.6412}{(49 \times 0.9917)} \cong 0.301 N$