Question:
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.
Answer:
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
Molar fraction of solvent = n(H2O)/(n(H2O) + n(solute))
Molar mass of solute = x g
Molar mass of solvent (H2O)=2×1+16 = 18g/mol
n1(H2O) = 90/18 = 5 mol , n2(H2O)=6 mol
n(solute)=30/x same n(solute)
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)
2.8= Po×(5x/(5x+30) …...(1)
2.9= Po×(6x/(6x+30)) …….(2)
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)
2.9(x+5)=2.8(x+6)
2.9x +14.5=2.8x +16.8
0.1x =2.3
X = 23 g
2.8= Po×(23×5/(23×5+30 ))
2.8=Po(115/145)
2.8×145/115 = Po
Po=3.53
Not what you are looking for? Go ahead and submit the question, we will get back to you.
