Volume $=1 \mathrm{~L}=1000 \mathrm{ml}$
Mass of solution $=1.54 \times 1000$
$=1540 \mathrm{~g}$
$\%$ purity of $\mathrm{H}_2 \mathrm{SO}_4$ is $70 \%$
So weight of $\mathrm{H}_3 \mathrm{PO}_4=0.7 \times 1540=1078 \mathrm{~g}$
Mole of $\mathrm{H}_3 \mathrm{PO}_4=\frac{1078}{98}=11$
Molarity $=\frac{11}{1 \mathrm{~L}}=11$
(આપેલ : $Mg$ નો પરમાણ્વીય દળ $24\, g\, mol ^{-1} ; N _{ A }=6.02 \times 10^{23}\, mol ^{-1}$ )
$\left[\right.$ આપેલ છે $: {N}_{{A}}=6.02 \times 10^{23}\, {~mol}^{-1}$ ,${Na}$નું આણ્વીય દળ $=23.0\, {u}]$