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 \(=\frac{11}{1 \mathrm{~L}}=11\)
[આણ્વિય દળ $: {Na}=23.0, {O}=16.0, {H}=1.0]$
$2{C_{57}}{H_{110}}{O_6}(s)\, + \,163\,{O_2}(g)\, \to \,114\,C{O_2}(g)\, + \,110\,{H_2}O(l)$