You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Relative Lowering of Vapour Pressure

0.6g urea is ...

Question

$0.6 g$ urea is added to $360 g$ water. Calculate lowering in vapor pressure for this solution
(Given: Vapour pressure of $H_{2} O$ is $35 m m$ of $H g$ )

0.027 m m

H g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.035 m m

H g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.017 m m

H g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.040 m m

H g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.017 m m$ of $H g$
Solution:- (C) $0.017 m m$ of $H g$
$\frac{P^{0} - P s}{P^{0}} = \frac{n}{n + N}$
Lowering in V.P.= $P^{0} \times \frac{n}{n + N}$
$= 35 \times \frac{\frac{0.6}{60}}{\frac{0.6}{60} + \frac{360}{18}} = 0.017 m m$ of Hg

Suggest Corrections

Similar questions

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. 18.2 g of urea is dissolved in 100g of water at

50^{o} C

. The lowering of vapour pressure produced is 5 mm Hg. Calculate the molecular mass of urea. The vapour pressure of water at

50^{o} C

is 92 mm Hg.

Q. The vapour pressure of a solution of methanol and ethanol at

20^{'} C

was found to be

70 m m

of Hg. Adding

10 g

of urea to

80 g

of this solution lowers the vapour pressure to

64.6 m m

of Hg. Determine composition of the original solution

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. 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:

Join BYJU'S Learning Program

0.6g urea is added to 360g water. Calculate lowering in vapor pressure for this solution (Given: Vapour pressure of H2O is 35mm of Hg)

The correct option is C 0.017mm of HgSolution:- (C) 0.017mm of HgP0−PsP0=nn+NLowering in V.P.= P0×nn+N =35×0.6600.660+36018=0.017mm of Hg

$0.6 g$ urea is added to $360 g$ water. Calculate lowering in vapor pressure for this solution
(Given: Vapour pressure of $H_{2} O$ is $35 m m$ of $H g$ )

The correct option is C $0.017 m m$ of $H g$
Solution:- (C) $0.017 m m$ of $H g$
$\frac{P^{0} - P s}{P^{0}} = \frac{n}{n + N}$
Lowering in V.P.= $P^{0} \times \frac{n}{n + N}$
$= 35 \times \frac{\frac{0.6}{60}}{\frac{0.6}{60} + \frac{360}{18}} = 0.017 m m$ of Hg