Question

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​

Answer 1

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:

Answer 2

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