Question

What is the freezing point of 10%(by weight) solution of ch3oh in water?

Answer 1

suppose the solution present is 100 g...then mass of methanol in it will be 10 g...

so mass of solvent (water)= 100 -10 = 90 g

molecular mass of methanol = CH3OH = 12 + 3X1 + 16 + 1 = 32 g/mole

as molality = number of moles of solute/mass of solvent in k.g.

so number of moles of methanol = 10/32 = 0.313

mass of solvent in k.g. = 90/1000 = 0.09 k.g.

so molality = 0.313/0.09 = 3.48

depression in freezing point = Kf X m

so depression in freezing point = 1.86 X 3.48 = 6.47

normal water freezes at 0 degree C ...here the depression in freezing point is 6.47,, means the freezing point has been lowered by 6.47 degree C.... so freezing point of solution = 0 -6.47 = -6.47 degree C

Answer 2

Explanation:

Tf=Kf×m

10% by weight means in 100 gms solvent 10 gms of solute is present

so molality=(10/32)/0.1

Kf must be given in Que