The mass of urea (Molar mass $60 g m o l^{- 1}$ ) in g that must be dissolved in $38 g$ of a solvent (having molar mass $57 g m o l^{- 1}$ ) , such that vapour pressure of the solution is reduced to $50 %$ of the vapour pressure of pure solvent is

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Solution

The correct option is B 40
Let, $P_{s} be the vapour pressure of solution$
$P^{0} be the vapour pressure of pure solvent$
$P_{s} = (\frac{50}{100}) P^{0} w_{s o l u t e} = ? M_{s o l u t e} = 60 g m o l^{- 1}, W_{s o l v e n t} = 38 g, M_{s o l v e n t} = 57 {g mol}^{- 1} ∵ \frac{P^{0} - P_{s}}{P_{s}} = \frac{w_{s o l u t e} \times M_{s o l v e n t}}{M_{s o l u t e} \times W_{s o l v e n t}} \Rightarrow \frac{P^{0} - (\frac{50}{100}) P^{0}}{(\frac{50}{100}) P^{0}} = \frac{w_{s o l u t e} \times 57}{60 \times 38}; \Rightarrow 1 = \frac{w_{s o l u t e}}{40} ∴ w_{s o l u t e} = 40 g$

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Similar questions

Q. The mass of urea (Molar mass

60 g m o l^{- 1}

) in g that must be dissolved in

38 g

of a solvent (having molar mass

57 g m o l^{- 1}

) , such that vapour pressure of the solution is reduced to

50 %

of the vapour pressure of pure solvent is

12 g

of non volatile solute having

(Molar mass = 120 g / m o l)

was dissolved in

9.9 m o l

of solvent. If the vapour pressure of pure solvent is

P_{A}^{\circ},

the vapour pressure of solution is:

6.0 g

of urea

(Molar mass = 60 g / m o l)

was dissolved in

9.9 m o l

of water. If the vapour pressure of pure water is

P_{A}^{\circ},

the vapour pressure of solution is:

QUESTION 2.36

100 g of liquid A (molar mass $140 g m o l^{- 1}$ ) was dissolved in 1000 g of liquid B (molar mass $180 g m o l^{- 1}$ ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.