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.

#### Solution

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

Now, the no. of moles of solvent (water), ` n_1 = (90g)/18 g mol^(-1) = 5 mol`

And, the no. of moles of solute, `n_2 = (30g)/(M mol^(-1)) = 30/M mol`

`p_1 = 2.8 kPa`

Applying the relation:

`(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)/((5M+30)/M)`

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

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

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

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

After the addition of 18 g of water

`n_1 = (90+18g)/18 = 6 mol`

`p_1 = 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/(6M+30) `

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

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

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

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

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

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

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

`=> 2.9 x 5 x (6m + 30) = 2.8 x 5 x (5M + 30)

=> 87M + 435 = 84M + 504

=> 3M = 69

=> M = 23u

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

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

`p_1^0/2.8 = (5xx23+30)/(5xx23)`

`=> 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.