The molarity of concentrated sulphuric acid $\left(ho=1.834 \mathrm{~g} \mathrm{~cm}^{-3}ight)$ containing $95 \%$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ by mass is

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Aicharya · Accepted Answer

Molarity $=\frac{95 \mathrm{~g} / 98 \mathrm{~g} \mathrm{~mol}^{-1}}{100 \mathrm{~g} / 1.834 \mathrm{~g} \mathrm{~cm}^{-3}}=0.01778 \mathrm{~mol} \mathrm{~cm}^{-3}=17.78 \mathrm{~mol} \mathrm{dm}^{-3}$

The molarity of concentrated sulphuric acid $\left(ho=1.834 \mathrm{~g} \mathrm{~cm}^{-3}ight)$ containing…

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