Question

3g of urea dissolved in 9 gram of water . calculate the relative lowering of vapour pressure​

Answer 1

Explanation:

n

B

=

60

3

=0.05;n

A

=

18

45

=2.5

Now,

p

0

△p

=x

B

=

2.5+0.05

0.05

=0.0196=0.02

Answer 2

Answer:

Relative lowering in vapour pressure is 0.090

Explanation:

Given: Mass of urea is 3g and Mass of water given is 9 gram.

To Find: Relative lowering of vapour pressure for the given solution.

Solution: When the solvent molecules that escape to vapour phase gets reduced , as a result the pressure exerted by vapour phase also gets reduced. This is known as relative lowering in vapour pressure.

The relative lowering in vapour pressure can be represented by formula as:

ΔP/P = Xb where;

ΔP is relative lowering in vapour pressure

P = Pure pressure and Xb = Mole fraction of solute

Xb = nb/na+nb where;

nb is mole fraction of solute and na is mole fraction of solvent

Moles = Given mass / molecular mass of substance

Mass of urea given is 3g and molecular mass of urea is 60g

Moles of urea (nb) = 3/60  = 0.05

Mass of water given is 9g and molecular mass of water is 18g

Moles of water (na) = 9/18 = 0.5

Mole fraction (Xb) = nb/nb+na = 0.05/0.5+0.05 = 0.05/0.55 = 0.090

So, ΔP/P = Xb where Xb is 0.090 so ΔP/P = 0.090

Relative lowering in vapour pressure is 0.090.

Code: #SPJ3