Question

The amount of solute (molar mass ) that must be added to 180g of water so that the vapour pressure of water is lowered by 10% is

Answer 1

Complete question:

The amount of solute (molar mass = 60 gm/mole ) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is

Complete Solution:

Given:

Mass of water, W = 180 gm

Lowering of vapour pressure = 10%

Molar Mass of solute = 60 gm

To Find:

The amount of solute that must be added for the required condition.

Calculation:

- Let the mass of solute be w gm.

- The number of moles of water, n1 = 180/18

⇒ n1 = 10

- The no of moles of solute, n2 = w/60

- We know that:

Relative lowering of vapour pressure = X(solute)

⇒ 0.1 = X

⇒ 0.1 = n2 / (n1 + n2)

⇒ (w/60) / {10 + (w/60)} = 0.1

⇒ w/60 = 0.1 × {(600 + w)/60}

⇒ w = 60 + 0.1 w

⇒ 0.9 w = 60

⇒ w = 60/0.9

⇒ w = 66.67 gm

- So, 66.67 gm of solute must be added for the required condition.

Answer 2

66.67g of solute should be added to the water to reduce its vapor pressure by 10%

Explanation:

Given: 180g of water

Find: Amount of solute to be added to reduce vapor pressure of water by 10%.

Note: The molar mass of solute is not mentioned. It must be 60 gm/mole.

Solution: Mass of water, W = 18 gm

Molar Mass of solute = 60 gm

Let the amount of solute required be X gm.

Number of moles of water in 180 g = n1 = 180/18 = 10

Number of moles of solute, n2 = X/60

Then the relative lowering of vapor pressure = P

Since P is 10%, we get:

  P = n2 / (n1+n2)

  10/100 = (X/60) (10 + x/60)

  10/100 = X / (600 + X)

  10 (600 + X) = 100X

  6000 + 10X = 100X

  90X = 6000

  X = 600 / 9

  X = 66.67 g

Hence 66.67g of solute should be added to the water to reduce its vapor pressure by 10%