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Question
The vapour pressure of pure water at a certain temperature is 23.80 mm Hg. If 1 mole of a non-volatile non-electrolytic solute is dissolved in 100g water, Calculate the resultant vapour pressure of the solution.
Solution
Relative lowering of vapour pressure = (P° - P) / P° = x2
`x_2 = "n"_2/"n"_1`
n2 = 0.1
n1 = `100/18`
`x_2 = 0.1/(5.55 + 0.1)`
`x_2 = 0.1/5.65`
`x_2 = 0.018`
P° = 23.8 mm Hg
Relative lowering of vapour pressure = `(23.80 - "P")/23.80 = 0.018`
23.80 - P = 0.428
P = 23.80 - 0.428
P = 23.37 mm Hg
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