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.)

MARKS Content Team · Accepted Answer

Correct answer: 1.36. 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, M=(number of moles)/volume =n/V =(( given mass of H_2 SO_4)/(molar mass) ) ( total mass of solution/ density ) M=(( 13 g)/(98 g/mol) ) ( 100 g/1.02 g/mL ) =(0.133 mol)/(98 mL) =(0.133 mol)/(0.098 L) M=1.357 mol/L ≃1.36

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.)

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