In sulphur estimation, 2.0 10-3 mol of an organic compound (X) (molar mass 76 g mol-1) gave 0.4813 g of barium sulphate (molar mass 233 g mol-1). The percentage of sulphur in the compound (X) is 10-1 % (Nearest integer)
Answer
435
Solution
Mass of X = n × M = 2.0 × 10-3 × 76 = 0.152 g
All S of X → BaSO4; 233 g BaSO4 contains 32 g S
⇒ mass of S = 32233 × 0.4813 = 0.06610 g
% S = 0.066100.152 × 100 = 43.49 % = 434.9 × 10-1 %
∴ ≈ 435
Answer: 435