A solution of non-volatile solute in water freezes at 0.3 degrees celsius the vapour pressure of pure water at 298 kelvin is it 23.51 mmhg and key app for water is 1.86 calculate the vapour pressure for this solution at 298 kelvin.
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it is IIT previous question paper
plz get me the answer it is useful for IIt and Sierra students.
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it is IIT previous question paper
plz get me the answer it is useful for IIt and Sierra students.
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Answer:
23.44mm Hg
Explanation:
∆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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