JEE MAIN - Chemistry (2022 - 30th June Morning Shift - No. 21)
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 ___________ $$\times$$ 10$$-$$1 g.
(Given molar mass of barium sulphate 233 g mol$$-$$1) (Nearest integer).
Answer
2
Explanation
$233 \mathrm{~g}\, \mathrm{BaSO}_{4}$ contains $=32 \mathrm{~g}$ sulphur
$0.752 \mathrm{~g}\, \mathrm{BaSO}_{4}$ contains $=\frac{32}{233} \times 0.752=0.1005 \mathrm{~g}$
Sulfur $\%=\frac{\text { Weight of Sulphur }}{\text { Weight of Compound }} \times 100$
$51.6=\frac{0.1005}{\text { Weight of compound }} \times 100$
$\Rightarrow$ Weight of compound $=\frac{0.1005}{51.6} \times 100=1.9 \times 10^{-1} \mathrm{~g}$
$$ \approx 2 \times 10^{-1} \mathrm{~g} $$
$0.752 \mathrm{~g}\, \mathrm{BaSO}_{4}$ contains $=\frac{32}{233} \times 0.752=0.1005 \mathrm{~g}$
Sulfur $\%=\frac{\text { Weight of Sulphur }}{\text { Weight of Compound }} \times 100$
$51.6=\frac{0.1005}{\text { Weight of compound }} \times 100$
$\Rightarrow$ Weight of compound $=\frac{0.1005}{51.6} \times 100=1.9 \times 10^{-1} \mathrm{~g}$
$$ \approx 2 \times 10^{-1} \mathrm{~g} $$
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