Question

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

[a.] the molecular mass of solute and

[b. ] vapour pressure of water at 298 K.


¤¤ Class 12th.​

Answer 1

The relative lowering of vapour pressure is given by the following expression

(p0solvent - psolution)/p0solvent  = n2/(n1 + n2)

Where p0solvent is the vapour prseure of the pure solvent, psolution is the vapour pressure of solution containing dissolved solute, n1 is the number of moles of solvent and n2 is the number of moles of solute.

For dilute solutions, n2 << n1, therefore the above expression reduces to

  (p0solvent - psolution)/ p0solvent = n2/n1

   = (w2 X M1) / (M2 X w1)  (i)

 

Where w1 and w2 are the masses and M1and M2 are the molar masses of solvent and solute respectively.

We are given that

   w2 = 30g

   w1 = 90g

  psolution = 2.8kpa

  p0solvent  = ? and  M2 = ?

Substituting these values in relation

(i), we get

(p0solvent -2.8)/ p0solvent = (30 X 18)/ (M2 X 90)

(p0solvent -2.8)/ p0solvent = 6/ M2  (1)

Similarly for second case we have the following values

  w2 = 30g

  w1 = (90 + 18)g = 108g

  psolution = 2.9 kpa

Therefore, we get

     (p0solvent -2.9)/ p0solvent = (30 X 18)/ (M2 X 108)

  = 5/M2  (2)

Dividing (1) by (2), We get

   (p0solvent -2.8)/ (p0solvent -2.9) = 6/5

Therefore,

  p0solvent = 3.4kpa

That is vapour pressure of water at 298K is 3.4 kpa.

  Substituting the value of p0solvent in (1), we get

 

(3.4- 2.8)/3.4 = 6/M2

  or  0.6/3.4 = 6/ M2

Therefore M2 = 34g

Therefore mass of solute is 34g and vapour pressure of water at 298K is 3.4 kpa.