Question

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

1. molar mass of the solute.
2. vapour pressure of water at 298 K.

Answer 1

(i) Let the molar mass of the solute be M g mol⁻¹.

Now, the number of moles of solvent (water), (n₁) = `(90  g)/(18  g  "mol"^-1)`

= 5 mol

And the number of moles of solute (n₂) = `(30  g)/(M  "mol"^-1) = 30/(M  "mol")`

p₁ = 2.8 kPa

By using Raoult’s law:

`(p_1^0 - p_1)/p_1^0 = n_2/(n_1 + n_2)`

⇒ `(p_1^0 - 2.8)/p_1^0 = (30/M)/(5 + 30/M)`

⇒ `1 - 2.8/p_1^0 = (30/M)/((5 M + 30)/M)`

⇒ `1 - 2.8/p_1^0 = 30/(5 M + 30)`

⇒ `2.8/p_1^0 = 1 - 30/(5 M + 30)`

⇒ `2.8/p_1^0 = (5 M + 30 - 30)/(5 M + 30)`

⇒ `p_1^0/2.8 = (5 M + 30)/(5 M)`    ...(i)

After the addition of 18 g of water

n₁ = `(90 + 18 g)/18` = 6 mol

p₁ = 2.9 kPa

Again, applying the relation,

`(p_1^0 - p_1)/p_1^0 = n_2/(n_1 + n_2)`

⇒ `(p_1^0 - 2.9)/p_1^0 = (30/M)/((6 + 30)/M)`

⇒ `1 - 2.9/p_1^0  = (30/M)/((6M + 30)/M)`

⇒ `1 - 2.9/p_1^0 = 30/(6 M + 30)`

⇒ `2.9/p_1^0 = 1 - 30/(6 M + 30)`

⇒ `2.9/p_1^0 = (6 M + 30 - 30)/(6 M + 30)`

⇒ `2.9/p_1^0 = (6 M)/(6 M + 30)`           

⇒ `p_1^0/2.9 = (6 M + 30)/(6 M)`   .......(ii)

Dividing equation (i) by (ii), we have:      

`2.9/2.8 = ((5 M + 30)/(5 M))/((6 M + 30)/(6 M))`

⇒ `2.9/2.8 xx (6 M + 30)/6 = (5 M + 30)/5`

⇒ 2.9 × 5 × (6m + 30) = 2.8 × 5 × (5M + 30)

⇒ 87M + 435 = 84M + 504

⇒ 3M = 69

⇒ M = 23 u

Therefore, the molar mass of the solute is 23 g mol⁻¹.

(ii) Putting the value of ‘M’ in equation (i), we have:

`p_1^0/2.8 = (5 xx 23 + 30)/(5 xx 23)`

`=> p_1^0/2.8   = 145/115`

`=> P_1^0 = 3.53`

Hence, the vapour pressure of water at 298 K is 3.53 kPa.