how much urea must be dissolved in 50 g of water so that the vapour pressure at room temperature is reduced by 25% ? also calculate the molality of the solution obtained
Answers
Answer:
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.
Explanation:
Calculation of amount of urea.
According to Raoult's Law , P∘A−PSPS=nBnA=WBMB×MAWA
Let P∘A=1matm,PS=0.75atmandP∘A−PS=1−0.75=0.25atm
(0.25atm)(0.75atm)=WB(60gmol−1)×(18gmol−1)(50g)
WB=0.250.75×6018×(50g)=55.55g
Calculation of molality of solution.
Molality of solution (m)=No. of moles of ureaMass of solvent (water)in kg=(55.55g)(60gmol−1)(501000kg)
55.5560×100050(molkg−1)=18.52molal