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.)

Answer 1

13% solution of Sulfuric acid consists of 13 g of $\text{H}_{\text{2}} \text{SO}_{\text{4}}$ and 87 g (100 g-13g) of water.

The molarity is,

$M =\frac{\text { number of moles }}{\text { volume }}=\frac{ n }{ V }=\frac{\left(\text { given mass of } H _{2} SO _{4} / \text { molar mass }\right)}{(\text { total mass of solution/density })}$

$M =\frac{\left({ }^{13 g / 98 g / mol }\right)}{(100 g / 1.02 g / mL )}=\frac{0.133 mol }{98 mL }=\frac{0.133 mol }{0.098 L }$

$M =1.357 \,mol / L \simeq 1.36$