Question

What is the molarity and normality of a $13\%$ solution (by weight) of sulphuric acid with a density of $1.02$ g/mL and also calculate the volume (in mL) that is required to prepare 1.5 $N$ of this acid to make it upto 100$ml$?

• A. Molarity$=1.35$, Normality$=2.70$, Volume$=180mL$
• B. Molarity$=2$, Normality$=4$, Volume$=200mL$
• C. Molarity$=1.5$, Normality$=3$, Volume$=150mL$
• D. None of these

Answer 1

The correct option is A Molarity$=1.35$, Normality$=2.70$, Volume$=180mL$

$\text{Molarity}=\frac{\text{percentage by weight}\times 10\times d}{Mw}$$=\frac{13\times 1.02\times 10}{98}=1.35M$

$13$% solution by weight means $13g$ of solute is dissolved in $87g$ of solvent.

$m=\frac{W_2\times 1000}{Mw\times W_1}$$=\frac{13}{98}\times \frac{1}{87}\times 1000=1.52m$

$Normality=Molarity\times 2$ ('n' factor for $H_{2}S{O}_4$)

$Normality=1.35\times 2=2.70$

For dilution,
$N_1{V}_1=N_2{V}_2$

or $100\times 2.70=1.5\times V_2$

or $V_2=180mL$