Practical Organic Chemistry and PurificationhardNUMERICAL

In a steam distillation process, an organic compound is distilled. The weight of water distilled aloPractical Organic Chemistry and Purification Chemistry Question

Question

In a steam distillation process, an organic compound $P$ is distilled. The weight of water distilled alongside is $18\ g$. If the vapour pressure of the pure organic compound at the distillation temperature is $100\ mm\ Hg$ and that of water is $200\ mm\ Hg$, calculate the mass (in grams) of the organic compound distilled. (Assume the molar mass of $P$ is $100\ g\ mol^{-1}$ and water is $18\ g\ mol^{-1}$).

Answer: 50

💡 Solution & Explanation

Using the steam distillation relation: $\frac{W_{water}}{W_P} = \frac{M_{water} \times P_{water}}{M_P \times P_P}$. Substituting the given values: $\frac{18}{W_P} = \frac{18 \times 200}{100 \times 100} = \frac{3600}{10000} = 0.36$. Thus, $W_P = \frac{18}{0.36} = 50\ g$.

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