A solution containing 30g of non volatile solute

Asked by admin @ 29/08/2021 in Chemistry viewed by 286 People

A solution containing 30g of non-volatile solute exactly in 90g of water has a vapour pressure of 2.8 kpa at 298 k. Furhter, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kpa at 298 k. Calculate : (i) molar mass of the solute (ii) vapour pressure of water at 298 k

Answered by admin @ 29/08/2021

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.

A solution containing 30g of non volatile solute

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