Verified Solution • 2025–26 Session moderate

QJEEM22Ch7b63: JEE Main 2022 Chemistry - Class 11 JEEM22Ch7b63 Solution: An organic compound with 51.6% sulfur is heated in a Carius…

An organic compound with 51.6% sulfur is heated in a Carius tube. The amount of this compound which will form 0.752 g of barium sulphate is ___________ × 10−1 g.

(Given molar mass of barium sulphate 233 g mol−1) (Nearest integer).

Step-by-Step Verified Solution

233 g BaSO4233 \mathrm{~g}\, \mathrm{BaSO}_{4} contains =32 g=32 \mathrm{~g} sulphur

0.752 g BaSO40.752 \mathrm{~g}\, \mathrm{BaSO}_{4} contains =32233×0.752=0.1005 g=\frac{32}{233} \times 0.752=0.1005 \mathrm{~g}

Sulfur %= Weight of Sulphur  Weight of Compound ×100\%=\frac{\text { Weight of Sulphur }}{\text { Weight of Compound }} \times 100

51.6=0.1005 Weight of compound ×10051.6=\frac{0.1005}{\text { Weight of compound }} \times 100

⇒\Rightarrow Weight of compound =0.100551.6×100=1.9×10−1 g=\frac{0.1005}{51.6} \times 100=1.9 \times 10^{-1} \mathrm{~g}

≈2×10−1 g

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