Steam distillation is carried out at a temperature where the sum of the partial vapour pressure of t — Practical Organic Chemistry and Purification Chemistry Question
Question
Steam distillation is carried out at a temperature where the sum of the partial vapour pressure of the organic liquid ($p_1$) and the partial vapour pressure of water ($p_2$) equals the atmospheric pressure. If the atmospheric pressure is $760\ mm\ Hg$ and the vapour pressure of the organic liquid is $45\ mm\ Hg$ at the distillation temperature, what is the vapour pressure of water ($p_2$) in $mm\ Hg$ at this temperature?
💡 Solution & Explanation
In steam distillation, the mixture boils when the sum of the partial vapour pressures of the organic liquid and water equals the atmospheric pressure: $P_{total} = p_1 + p_2$. Thus, $760 = 45 + p_2 \implies p_2 = 760 - 45 = 715\ mm\ Hg$.