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Question 19

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    n/ (n+ 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/p1= 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    n/ (n+ 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/p1= 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

⇒  p1=  (145 x 2.8) / 115

⇒  p1=  3.53

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

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