JEE MAIN - Chemistry (2024 - 31st January Evening Shift - No. 29)
The molarity of $$1 \mathrm{~L}$$ orthophosphoric acid $$\left(\mathrm{H}_3 \mathrm{PO}_4\right)$$ having $$70 \%$$ purity by weight (specific gravity $$1.54 \mathrm{~g} \mathrm{~cm}^{-3}$$) is __________ $$\mathrm{M}$$.
(Molar mass of $$\mathrm{H}_3 \mathrm{PO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}$$)
Answer
11
Explanation
Specific gravity (density) $$=1.54 \mathrm{~g} / \mathrm{cc}$$.
Volume $$=1 \mathrm{~L}=1000 \mathrm{~ml}$$
Mass of solution $$=1.54 \times 1000$$
$$=1540 \mathrm{~g}$$
$$\%$$ 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 $$=\frac{11}{1 \mathrm{~L}}=11$$
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