$3 g$ urea is dissolved in $45 g$ of water. The relative lowering of vapour pressure is:

0.05

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

0.04

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

0.02

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

0.01

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

Open in App

Solution

The correct option is B $0.02$
$n_{B} = \frac{3}{60} = 0.05; n_{A} = \frac{45}{18} = 2.5$

Now,

$\frac{△ p}{p_{0}} = x_{B} = \frac{0.05}{2.5 + 0.05} = 0.0196 = 0.02$

Suggest Corrections

Similar questions

Q. 6gm. of urea is dissolved in 90 gm of water. The Relative lowering of vapour pressure is equal to

___

3 g

urea is dissolved in

45 g

of water. The relative lowering of vapour pressure is :

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. 6gm of urea is dissolved in 90gm of water. Relative lowering of vapour pressure is:

When 30 g of urea are dissolved in 9.0 g of water, the relative lowering of vapour pressure is:

Join BYJU'S Learning Program

3g urea is dissolved in 45g of water. The relative lowering of vapour pressure is:

The correct option is B 0.02nB=360=0.05;nA=4518=2.5Now,△pp0=xB=0.052.5+0.05=0.0196=0.02

$3 g$ urea is dissolved in $45 g$ of water. The relative lowering of vapour pressure is:

The correct option is B $0.02$
$n_{B} = \frac{3}{60} = 0.05; n_{A} = \frac{45}{18} = 2.5$

Now,

$\frac{△ p}{p_{0}} = x_{B} = \frac{0.05}{2.5 + 0.05} = 0.0196 = 0.02$