A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO 2 ceases. The volume of CO 2 at 750 mm Hg pressure and at 298 K is measured to be 123.9 mL. A 1.5 g of the same sample requires 150 mL of (M/10) HCl for complete neutralization. Calculate the percentage composition of Na 2 SO 4 in the original mixture.

Step-by-step solution

In 2g sample → CO 2 Liberated Moles of CO 2 = 7 5 0 7 6 0 × 1 2 3 . 9 1 0 0 0 × 1 0 . 0 8 2 × 1 2 9 8 = 5 × 1 0 - 3 moles = 5 mM ≡ 1 0 mM of NaHCO 3 in mixture Moles of NaHCO 3 in 2g sample = 2 x 5 x 10 mM 1.5g sample ≡ 150 × 1 1 0 = 15 mM of HCl 2.0g sample ≡ 15 × 2 1 · 5 = 20 mM · 20 mM HCl will be utilised by 10 mM NaHCO 3 + 5 mM Na 2 CO 3 present in 2g sample NaHCO 3 = 10 mM = 10 × 1 0 - 3 × 8 4 = 0.84g 42% Na 2 CO 3 = 5 mM = 5 × 1 0 - 3 × 1 0 6 = 0.53g = 26.5% Na 2 SO 4 = (By difference) = 0.63g 31.5%