$30 g$ of urea $(M = 60 g m o l^{- 1})$ is dissolved in $846 g$ of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at $298 K$ is $23.8 m m H g$ .

Open in App

Solution

According to Raoult's law:

$P_{s o l u t i o n} = P_{w a t e r}^{o} X_{w a t e r} = 23.8 \times \frac{47}{49} = 22.83 m m H g$

Note:
$X_{w a t e r} = \frac{n_{w a t e r}}{n_{u r e a} + n_{w a t e r}} = \frac{846 / 18}{30 / 60 + 846 / 18} = \frac{47}{2 + 47} = \frac{47}{49}$

Suggest Corrections

Similar questions

Q. Vapour pressure of pure water at

298 K

23.8 m m H g

50 g

of urea

(N H_{2} C O N H_{2})

is dissolved in

850 g

of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Q. (a) 30 g of urea (m = 60 g

m o l^{- 1}

) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences betwen ideal solutions and non- ideal solutions.

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Q. Vapour pressure of pure water at

298 K

23.8 m m

H g

50 g

of urea is dissolved in

850 g

of water. The vapour pressure of water for this solution and its relative lowering are respectively:

Q. Vapor pressure of pure water at

298 K

23.8 m m H g

50 g

of urea is dissolved in

850 g

of water. Calculate the vapor pressure of water for this solution and it's relative lowering.

Join BYJU'S Learning Program

30 g of urea (M=60 g mol−1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.

According to Raoult's law:Psolution=PowaterXwater=23.8×4749=22.83 mm HgNote: Xwater=nwaternurea+nwater=846/1830/60+846/18=472+47=4749

$30 g$ of urea $(M = 60 g m o l^{- 1})$ is dissolved in $846 g$ of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at $298 K$ is $23.8 m m H g$ .

According to Raoult's law:

$P_{s o l u t i o n} = P_{w a t e r}^{o} X_{w a t e r} = 23.8 \times \frac{47}{49} = 22.83 m m H g$

Note:
$X_{w a t e r} = \frac{n_{w a t e r}}{n_{u r e a} + n_{w a t e r}} = \frac{846 / 18}{30 / 60 + 846 / 18} = \frac{47}{2 + 47} = \frac{47}{49}$