At room temperature, a dilute solution of urea is prepared by dissolving 40.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, what will be the lowering of vapour pressure? (molar mass of urea = 60 g mol-1)

Question

At room temperature, a dilute solution of urea is prepared by dissolving 40.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, what will be the lowering of vapour pressure? (molar mass of urea = 60 g mol-1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Relative lowering of vapour pressure, is given
by,

\frac{P ^{o} - P}{P ^{o}} = x_{A} = \frac{n _{A}}{n _{A} + n _{B}} ≃ \frac{n _{A}}{n _{B}}

Given,

P^{o} = 35 mm H g, n_{urea} = \frac{0.60}{60}

n_{water} = \frac{360}{18}

\frac{P ^{o} - P}{35} = \frac{0.6 \times 18}{60 \times 360} = \frac{1}{2000}

Δ P = P^{o} - P = 0.017

Q. At room temperature, a dilute solution of urea is prepared by dissolving 40.60g of urea in 360g of water. If the vapour pressure of pure water at this temperature is 35mmHg, what will be the lowering of vapour pressure? (molar mass of urea =60gmol−1)

Relative lowering of vapour pressure, is given by, PoPo−P​=xA​=nA​+nB​nA​​≃nB​nA​​ Given, Po=35mmHg,nurea​=600.60​ nwater​=18360​ 35Po−P​=60×3600.6×18​=20001​ ΔP=Po−P=0.017

Q. At room temperature, a dilute solution of urea is prepared by dissolving $40.60 g$ of urea in $360 g$ of water. If the vapour pressure of pure water at this temperature is $35 mm H g$ , what will be the lowering of vapour pressure? (molar mass of urea $= 60 g m o l^{- 1}$ )