JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Practical and Analytical Chemistry

Every sulphur atom of the sample is precipitated as barium sulphate, so the sulphur present is the mass of barium sulphate scaled by $32/233$, the ratio of the atomic mass of sulphur to the molar mass of barium sulphate. This gives $\frac{32}{233}\times 1.4439=0.1983$ g of sulphur. As a fraction of the $0.471$ g sample, $\frac{0.1983}{0.471}\times 100=42.1$. Rounded to the nearest integer the sulphur content of the compound is $42\%$.