In sulphur estimation,$0.157 \ g$ of an organic compound gave $0.4813 \ g$ of barium sulphate. What is the percentage of sulphur in the compound?

(42.10%) The molecular mass of $BaSO_4 = 137 + 32 + 4 \times 16 = 233 \ g/mol$.
The amount of sulphur in $233 \ g$ of $BaSO_4$ is $32 \ g$.
The amount of sulphur in $0.4813 \ g$ of $BaSO_4$ is $\frac{32 \times 0.4813}{233} \ g$.
The percentage of sulphur in the organic compound is calculated as:
$\text{Percentage of sulphur} = \frac{32 \times 0.4813 \times 100}{233 \times 0.157} \%$.
$\text{Percentage of sulphur} = 42.10 \%$.