How much urea should be dissolved in 50 gram of water is so that the vapour pressure at room temperature is reduced by 25% calculate the molality of solution obtained?
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Answered by
83
The percentage reduction in the vapor pressure is given by the relation,
ΔP/P = W₂M₁/W₁M₂
where,
W₁ = weight of water = 50 g
W₂ = weight of urea
M₁ = molar mass of water = 18
M₂ = molar mass of urea = 60
=> 0.25 = (W₂ x 18)/(50x60)
=> 18W₂ = 750
=> W₂ = 750/18 = 41.67 grams
molality of the solution
m = moles of urea/weight of water in kg
moles of urea = W₂/M₂ = 41.67/60 = 0.7
weight of water in kg = 50/1000 = 0.05
=> molality = 0.7/0.05 = 14
Hence weight of urea should be 41.67 grams and molality of the solution will be 14.
Answered by
27
Answer: mass of urea :- 55.55 gram
Molality:- 18.518m
Explanation:- please look at the attached pic
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