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a solution containing 30 g of non volatile solute...

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:

Let, the molar mass of the solute be M g mol ^{- 1}

Now, the no. of moles of solvent (water),n₁= 90g / 18g mol^-1

And, the no. of moles of solute,n₂= 30g / M mol^-1 = 30 / M mol

p₁= 2.8 kPa

Applying the relation:

(p₁⁰ - p₁) / p₁⁰= n₂/ (n₁+ n₂)

⇒ (p₁⁰ - 2.8) / p₁⁰= (30/M)/ {5+ (30/M)}

⇒ 1 - (2.8/p₁⁰) = (30/M) / {(5M+30)/M}

⇒ 1 - (2.8/p₁⁰) = 30 / (5M + 30)

⇒ 2.8/p₁⁰= 1 - 30 / (5M + 30)

⇒ 2.8/p₁⁰ = (5M + 30 - 30) / (5M + 30)

⇒ 2.8/p₁⁰= 5M / (5M+30)

⇒ p₁⁰ / 2.8 = (5M+30) / 5M ----------------(1)

After the addition of 18 g of water:

n₁ = (90+18g) / 18 = 6 mol

and the new vapour pressure is p₁ = 2.9 kPa (Given)

Again, applying the relation:

(p₁⁰ - p₁) / p₁⁰= n₂/ (n₁+ n₂)

⇒ (p₁⁰ - 2.9) / p₁⁰= (30/M)/ {6+ (30/M)}

⇒ 1 - (2.9/p₁⁰) = (30/M) / {(6M+30)/M}

⇒ 1 - (2.9/p₁⁰) = 30 / (6M + 30)

⇒ 2.9/p₁⁰= 1 - 30 / (6M + 30)

⇒ 2.9/p₁⁰ = (6M + 30 - 30) / (6M + 30)

⇒ 2.9/p₁⁰= 6M / (6M+30)

⇒ p₁⁰ / 2.9 = (6M+30) / 6M ----------------(2)

Dividing equation (1) by (2),we get:

2.9 / 2.8 = {(5M+30) / 5M} / {(6M+30) / 6M}

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

⇒ 2.9 x (6M +30) x 5 = (5M+30) x 2.8 x 6

⇒ 87M + 435 = 84M + 504

⇒ 3M = 69

⇒ M = 23u

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

(ii) Putting the value of 'M' in equation (i), we get:

⇒ p₁⁰ / 2.8 = (5M+30) / 5M

⇒ p₁⁰ / 2.8 = (5x23+30) / 5x23

⇒ p₁⁰= (145 x 2.8) / 115

⇒ p₁⁰= 3.53

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

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