A bottle of${\mathrm{H}}^3 {\mathrm{PO}}^4$ solution contains 70% (w'w) acid. If the density of the solution is $1.54 \mathrm{g} {\mathrm{cm}}^{-3}$, the volume of the ${\mathrm{H}}^{3^{ }}{\mathrm{PO}}^4$ solution required to prepare 1 L of 1 N solution is :

${\mathrm{H}}_3{\mathrm{PO}}_3$ solution contains 70% w/w acid

d = 1.54 gm/ml

$70\mathrm{gm} {\mathrm{H}}_3{\mathrm{PO}}_4\mathrm{\_\_\_\_\_}$ in 100 gm solution

$\begin{array}{l}d=\frac{M}{V}1.54=\frac{100}{V}V=\frac{100}{1.54}\\ N_1{V}_1=N_2{V}_2\\ 3\times V_1\times \frac{70}{98}\times \frac{1000}{100}\times 1.54=1\times 1000\\ V_1=\frac{98000\times 100}{70\times 1.54\times 1000\times 3}\\ V_1=30\mathrm{ml}\end{array}$