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Chemistry · Colligative properties of dilute solutions: a relative lowering of vapour pressure, depression of freezing point, the elevation of boiling point and osmotic pressure

A mixture of chlorobenzene and wate (immiscible) boils at at an external pressur

A mixture of chlorobenzene and wate (immiscible) boils at \( 90.3^{\circ} \mathrm{C} \) at an external pressure of \( 740.2 \mathrm{mm} . \) The vapour pressure of pure water at \( 90.3^{\circ} \mathrm{C} \) is \( 530.1 \mathrm{mm} \). Calculate the \( \% \) composition of distillate:

  • A. \( H_{2} O=35 \% \)
  • B. \( H_{2} O= \) 22\%
  • C. \( H_{2} O= \) २९\% \( \%= \)
  • D. \( H_{2} O=71 \% \)

Step-by-step solution

For immiscible liquids, total pressure equals sum of vapor pressures: P_total = P_water + P_chlorobenzene. Given P_total = 740.2 mm and P_water = 530.1 mm, P_chlorobenzene = 210.1 mm. Moles in vapor are proportional to vapor pressures: n_water/n_chlorobenzene = P_water/P_chlorobenzene = 530.1/210.1 ≈ 2.523. Mass ratio = (n_water × M_water)/(n_chlorobenzene × M_chlorobenzene) = 2.523 × (18.015/112.56) ≈ 0.4038. Mass fraction of water = 0.4038/(1+0.4038) ≈ 0.2877 or 28.8%, closest to 29%.
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