Question

(a) What is the normality of a $96\%$ solution of $H_{2}S{O}_4$ of specific gravity $1.84$?
(b) How many $mL$ of $96\%$ sulphuric acid solution is necessary to prepare one litre $0.1N{H}_{2}S{O}_4$?
(c) To what volume should $10mL$ of $96\%$ $H_{2}S{O}_4$ be diluted to prepare $2N$ solution?

Answer 1

Mass of $1litre$ of $H_{2}S{O}_4$ solution
$=Vol.\times Density$
$=1000\times 1.84=1840g$
Mass of $H_{2}S{O}_4$ present in one litre $96$% $H_{2}S{O}_4$ solution
$=\frac{96}{100}\times 1840=1766.4g$
Strength of $H_{2}S{O}_4$ solution $=1766.4g/L$
(a) $Normality=\frac{Strength}{Eq.mass}=\frac{1766.4}{49}=36.05N$
(b) Let the volume taken be $V_{1}mL$
Applying $N_1{V}_1=N_2{V}_2$N_{1} = 36.05\ N, V_{1} = ?, N_{2} = \dfrac {N}{10}, V_{2} = 1000\ mL36.05\times V_{1} = \dfrac {1}{10}\times 1000$So,$V_{1} = \dfrac {1000}{36.05\times 10} = 2.77\ mL$i.e.,$2.77\ mL$of$H_{2}SO_{4}$isdilutedtoonelitre.\left(c\right)$\underset {Before\ dilution}{N_{B}V_{B}} = \underset {After\ dilution}{N_{A}V_{A}}10\times 36.05 = V_{A} \times 2V_{A} = 180.25\ mL$i.e.,$10\ mL$ofgiven$H_{2}SO_{4}$isdilutedto$180.25\ mL$.