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 : (i) molar mass of the solute (ii) vapour pressure of water at 298 K.

Explanation

(i) First Experiment : Suppose the molar mass of the solute = Mg `"mol"^(-1 )`
Number of moles, `n_2` (solute) = 30/M moles
Number of moles, `n_1` (solvent, `H_2O`) = `(90g)/(10g "mol"^(-1) ) = `5 moles
Using Raoult.s law equation and substituting values, we have
`(p^0 -p_s)/(p^0) = (n_2)/(n_1 + n_2) i.e., (p^0 -2.8)/(p^0) = (30//M)/(5 + 30//M)`
` 1- (2.8)/(p^0) = (30 //M)/(5 + 30 //M)`
`(2.8)/(p^0) = 1 - (30 //M)/(5 + 30//M) = (5 + 30//M -30//M)/(5 + 30 //M) = (5)/(5 + 30 // M)`
` (p^0)(2.8) = (5 + 30//M)/(5) = 1 + (6)/(M)` ..(1)
Second Experiment :
After adding 18 g of water.
Number of moles of water `n_1` = 6 moles
Substituting values in Raoult.s law equation, we have
`(p^0 - 2.9)/(p^0) = (30//M)/(6+ 30//M)`
`1-(2.9)/(p^0) = (30 //M)/(6 + 30//M)`
`(2.9)/(p^0) = 1 - (30 //M)/(6 + 30//M) = (6+ 30//M - 30 //M)/(6 + 30 //M) = (6)/(6+ 30//M)`
`(p^0)/(2.9) = (6 + 30//M)/(6) = 1 + 5/M` ....(2)
There are two unknown quantities and two equations.
Dividing equation (1) by equation (2), we get
`(2.9)/(2.8) = (1+ 6//M)/(1 + 5//M) " or " 2.9 (1 + 5/M)= 2.8 (1+ 6/M)`
`2.9 + (14.5)/(M) = 2.8 + (16.8)/(M) " or " (2.3)/(M) = 0.1 " or " M = 23 g "mol"^(-1)`
(ii) Substituting M = 23 in equation (2), we get
`(p^0)/(2.8) = 1 + 6/23 = 29/23`
`p^0 = 29/23 xx 2.8 = 3.53 kPa`