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.) (A)  (B)  (C)  (D)

Tardigrade Admin · 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,

M = \frac{number of moles}{volume} = \frac{n}{V} = \frac{( given mass of H _{2} S O _{4} / molar mass )}{( total mass of solution/density )}

M = \frac{( ^{13 g /98 g / m o l} )}{( 100 g /1.02 g / m L )} = \frac{0.133 m o l}{98 m L} = \frac{0.133 m o l}{0.098 L}

M = 1.357 m o l / L ≃ 1.36

Q. 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.)

Q. 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.)