Question

Calculate the molarity (C_M) of a solution containing 33.5g Na₂SO₄ in a final volume of 237mL.

The answer is supposed to be 0.995M

Answer 1

Mass of Na₂SO₄ = 33.5 g

Volume of solution = 237 mL. ( 1mL = 10^-3 L )

Volume of solution = 237* 10^-3 L

Volume of solution = 0.237 L

Atomic mass of Na = 23 , S= 32 and O = 16 u

Molar mass of Na₂SO₄ = 2*( Atomic mass of Na) + ( Atomic mass of S) + 4*( Atomic mass of O )

Molar mass of Na₂SO₄ = 2*23 + 32 + 4*16 g /mol

Molar mass of Na₂SO₄ = 46 + 32 + 64 g/ mol

Molar mass of Na₂SO₄ = 142 g / mol

Moles of Na₂SO₄ = ( Mass of Na₂SO₄) / ( Molar mass of Na₂SO₄)

Moles of Na₂SO₄ = 33.5 / 142 = 0.235915 mol

Molarity = ( Moles of Na₂SO₄ ) / ( Volume of solution in L)

Molarity = 0.235915/ 0.237 M

Molarity = 0.99542 M