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.

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.

Solution :

Step-1  What is given?

Mass  of non volatile solute     = 30 g

Mass of solvent (w1)                = 90 g

Pressure of  solution (p1)         = 2.8 kpa

Mass  of solute is  same                    = 30 g

Mass of solvent(w2)                 = 90+18 = 108 g

Pressure of solution (p2)          = 2.9 kpa

Pressure of solution = (P^o)× molar fraction of solvent                        ... (1)

Molar fraction of solvent = n_(H2O)/(n_(H2O) + n_(solute))                          ... (2)

                                                          ... (3)

Molar mass of solute = x g

Molar mass of solvent (H2O)=2×1+16 = 18g/mol

Plug the value in  equation (3)

n¹_(H2O) = 90/18 = 5 mol , n²_(H2O)=6 mol

n_(solute)=30/x    same n_(solute)

plug in equation (2) we get

molar fraction of first solution  X¹_H2O = 5/(5+30/x)=5x/(5x+30)

molar fraction of second solution X²_H2O = 6/(6+30/x)=6x/(6x+30)

plug these value in equation (1) we get

2.8= P^o×(5x/(5x+30)       …...(5)

2.9= P^o×(6x/(6x+30))      …….(6)

Now divide equation second by first we get

2.9/2.8= (P⁰×(6x/(6x+30)))/ (P^o×(5x/(5x+30)))

2.9/2.8=6x(5x+30)/5x(6x+30)

2.9/2.8=(x+6)/(x+5)

Now cross multiply we get

2.9(x+5)=2.8(x+6)

2.9x +14.5=2.8x +16.8

0.1x =2.3

X = 23 g

Now plug the value in equation (5) we get

2.8= P^o×(23×5/(23×5+30 ))

2.8=P^o(115/145)

2.8×145/115 = P^o

3.53= P^o