Question

A solution is made by dissolving 30g of a non volatile solute in 90 g of water. It has a vapour pressure of 2.8kpa at 298k. The vapour pressure of pure water is 3.46 kpa.Calculate the molar mass of the solute

Answer 1

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 = (Po)× 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 X1H2O = 5/(5+30/x)=5x/(5x+30)

molar fraction of second solution X2H2O = 6/(6+30/x)=6x/(6x+30)

plug these value in equation (1) we get

2.8= Po×(5x/(5x+30) …...(5)

2.9= Po×(6x/(6x+30)) …….(6)

Now divide equation second by first we get

2.9/2.8= (P0×(6x/(6x+30)))/ (Po×(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= Po×(23×5/(23×5+30 ))

2.8=Po(115/145)

2.8×145/115 = Po

3.53= Po

Answer 2

Answer:

34g.

Explanation:

plz mark as brainliest