QJEEM24Ch55e4: JEE Main 2024 Chemistry - Class 11 JEEM24Ch55e4 Solution: The molarity of 1   L orthophosphoric acid ( H 3 PO 4 )…

Specific gravity (density) $$=1.54\mathrm{g}/\mathrm{cc}$$.

Volume $$=1\mathrm{L}=1000\mathrm{ml}$$

Mass of solution $$=1.54\times 1000$$

$$=1540\mathrm{g}$$

$$\mathrm{\%}$$ purity of $${\mathrm{H}}_2{\mathrm{SO}}_4$$ is $$70\mathrm{\%}$$

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 $$=\frac{11}{1\mathrm{L}}=11$$