Skip to content
dronacharya.ai
← JEE Main 2026 · 4 Apr · Shift 2
ChemistryInteger+4 / -1Question 75 of 75

JEE Main 2026 · 4 Apr · Shift 2, Chemistry question 75

In sulphur estimation, 2.0×10−32.0\times10^{-3} mol of an organic compound (X) (molar mass 76 g mol−1^{-1}) gave 0.4813 g of barium sulphate (molar mass 233 g mol−1^{-1}). The percentage of sulphur in the compound (X) is ________ ×10−1\times10^{-1} % (Nearest integer)

Answer and worked solution

Official answer: 435

Sulphur. 0.4813233=2.066×10−3\frac{0.4813}{233}=2.066\times10^{-3} mol S, i.e. 0.06610.0661 g.

Compound. 2.0×10−3×76=0.1522.0\times10^{-3}\times76=0.152 g.

Percentage. 0.06610.152=43.5%=435×10−1%\frac{0.0661}{0.152}=43.5\%=435\times10^{-1}\%.

Answer: 435

Topics:Qualitative and quantitative analysis of organic compounds
← Previous question