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A solution of a nonvolatile solute in water freezes at – 0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and Kf for water is 1.86 K kg mol-1. Calculate the vapour pressure of this solution at 298 K.
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A solution of a nonvolatile solute in water freezes at – 0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and Kf for water is 1.86 K kg mol-1. Calculate the vapour pressure of this solution at 298 K.
4 years ago
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Hello Student,
Please find the answer to your question
∆Tf = Kf.m
Po – p/po = moles of solute/moles of solvent
Depression in freezing point, ∆Tf = Kf m
∴ m ∆Tf/Kf = 0.30/1.86 = 0.161
According to Raoult’s law
Po – p/po = No. of moles of solute/No. of moles of solvent
23.51 – p/23.51 = 0.161/1000/ 18 = 0.161 *18/1000
(∵ No. of moles of H2O = 1000/18)
On usual calculations,
23.51 –p/23.51 = 0020898
P = 23.51 – 23.51 * 0.0020898 = 23.51 – 068
p = 23.44 mm Hg
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