The molarity (in mol/liter) of a 13% solution (by weight) of sulphuric acid with a density of 1.02 g/mL is
(Note: Report your answer in the required format by relevant rounding up.)

Question

The molarity (in mol/liter) of a 13% solution (by weight) of sulphuric acid with a density of 1.02 g/mL is (Note: Report your answer in the required format by relevant rounding up.)

TestHub · Accepted Answer

13% solution of Sulfuric acid consists of 13 g of $H_{2} {SO}_{4}$ and 87 g (100 g-13g) of water.
The molarity is,
$\begin{array}{l} M = \frac{number of moles}{volume} = \frac{n}{V} = \frac{(\frac{{given mass of H}_{2} {SO}_{4}}{molar mass})}{(total mass of solution / density)} \\ M = \frac{(\frac{13 g}{98 g/mol})}{(100 g/1 .02 g/mL)} = \frac{0.133 mol}{98 mL} = \frac{0.133 mol}{0.098 L} \\ M = 1.357 mol/L ≃ 1.36 \end{array}$

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