Question

1. what is the molar mass of X(i = 1) if 50.00 water containing 2.50 g of X has a freezing point of -11.5 degree Celcius

2.Using information from Question 1, recalculate the molar mass of X if i = 2

Answer 1

normal boiling point of water solution = 0 C

observed boiling point of water solution = -11.5 C

depression in freezing poin dTf = 0 - ( -11.5 )

dTf = 11.5 C

we know that

depression in freezing point (dTf) = i x Kf x m

for solvent water Kf = 1.86

11.5 = 1 x 1.86 x m

m = 6.18

now

molality = moles of solute / mass of solvent (kg)

6.18 = moles of X / 0.05

moles of X = 0.309

now

molar mass = mass /moles

molar mass of X = 2.5 / 0.309

molar mas of X = 8.08 g

2)

normal boiling point of water solution = 0 C

observed boiling point of water solution = -11.5 C

depression in freezing poin dTf = 0 - ( -11.5 )

dTf = 11.5 C

we know that

depression in freezing point (dTf) = i x Kf x m

for solvent water Kf = 1.86

11.5 = 2 x 1.86 x m

m = 3.09

now

molality = moles of solute / mass of solvent (kg)

3.09 = moles of X / 0.05

moles of X = 0.1545

now

molar mass = mass /moles

molar mass of X = 2.5 / 0.1545

molar mas of X = 16.17 g