MCQ
When $0.01\,mol$ of an organic compound containing $60\,\%$ carbon was burnt completely, $4.4\,g$ of $CO _2$ was produced. The molar mass of compound is $.........g\,mol ^{-1}$ (Nearest integer)
- A$100$
- B$50$
- ✓$200$
- D$150$
$\text { mass of carbon }=0.01 M \times \frac{60}{100}$
$\text { moles of carbon }=\frac{0.01 M }{12} \times \frac{60}{100}$
moles of $CO _2$ from combustion $=\frac{4.4}{44}=$ moles of carbon
$\frac{0.01\,M }{12} \times \frac{60}{100}=\frac{4.4}{44}$
$M =\frac{4.4}{44} \times \frac{100}{60} \times \frac{12}{0.01}=200\, gm / mol$
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