Question

Mass of a non-volatile solute (molar mass = 45 g
mol-?) that should be dissolved in 90 g water to
reduce its vapour pressure to 75% is

Answer 1

Answer:

The mass of a non-volatile solute of molar mass 40 g /mol that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is 10 g.

Let w g of the non-volatile solute is used. Its molar mass is 40 g/mol.

The number of moles of non volatile solute =

40

w

​

=0.025w moles.

114 g of octane (molar mass 114 g/mol) corresponds to

114

114

​

=1 mole.

The mole fraction of non volatile solute is =

1+0.025w

0.025w

​

. The mole fraction of non volatile solute is proportional to the relative lowering in the vapour pressure.

P

0

P

0

−P

​

=X

B

​

100

20

​

=

1+0.025 w

0.025w

​

100(0.025w)=20(1+0.025w)

2.5w=20+0.50w

2w=20

w=10 g

Hence, the correct answer is 10g

Explanation:

Answer 2

Given:

Mass of a non-volatile solute (molar mass = 45 g  mol-?) that should be dissolved in 90 g water to  reduce its vapour pressure to 75% is

To find:

Mass of a non-volatile solute

Solution:

Let x g of the non-volatile solute is used. Its molar mass is 45 g/mol.

The number of moles of non-volatile solute = x/45 = 0.022x

90 g of water (molar mass 18 g/mol) = 90/18 = 5 moles

The mole fraction of non volatile solute is = 0.022x/(1+0.022x)

The mole fraction of non volatile solute is proportional to the relative lowering in the vapour pressure.

75% = 75/100 = 0.022x/(1+0.022x)

$\dfrac{75}{100} = \dfrac{0.022x}{5+0.022x}$

75(5 + 0.022x) = 0.022x(100)

375 + 1.65x = 2.2x

375 = 0.55x

x = 375/0.55 = 681.81

∴ Mass of a non-volatile solute 681.81 g