Question

23 Derive an Expression to find out
the molar mass of solute by relative
lowering of vapower pressusce as a
Colligative property​

Answer 1

Answer:

The expression for the relative lowering of vapour pressure is

P

0

ΔP

=

P

0

P

0

−P

=X

2

.....(1)

Here,

P

0

ΔP

and

P

0

P

0

−P

represents relative lowering of vapour pressure and X

2

represents mole fraction of solute.

But the mole fraction of solute

X

2

=

n

1

+n

2

n

2

=

M

2

W

2

+

M

1

W

1

M

2

W

2

.....(2)

Here, W

1

and M

1

are the mass and molar masses of solvent and W

2

and M

2

are the mass and molar masses of solute. n

1

and n

2

are the number of moles of solute and solvent respectively.

For dilute solutions n

1

>>n

2

So n

2

may be neglected in comparison with n

1

.

Hence, equation (2) becomes

X

2

=

n

1

n

2

=

M

1

W

1

M

2

W

2

.....(3)

From equation (1) and equation (3)

P

0

ΔP

=

P

0

P

0

−P

=

M

1

W

1

M

2

W

2

This gives the relationship between relative lowering of vapour pressure and molar mass of solute.

Explanation:

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