Question

When 1.31 g of an unknown non-electrolyte is dissolved in 50.0 g of acetone, the freezing point decreased to -94.1 degrees C from -93.4 degrees C. If the Kfp of the solvent is 2.4 K/m, calculate the molar mass of the unknown solute

Answer 1

Given,

mass of unknown solute = 1.31 g

mass of solvent acetone = 50 g = 0.050 kg

molality = moles of solute/ kg of solvent

i = 1 [Van't hoff factor]

Kf = 2.4 oC/m

dTf = -94.1 - (-93.4) = -0.7 oC

Using,

dTf -i.Kf.m

therefore

-0.7 = -1 x 2.4 x m

m = 0.292

and molality = number of moles / solvent in kg

number of moles = 0.292 mole / kg *0.050 kg

= 0.0146 moles

Molar mass = amount in g / number of moles

= 1.31g /0.0146 mole

= 89.73 g/ mole

Given,

mass of unknown solute = 1.31 g

mass of solvent acetone = 50 g = 0.050 kg

molality = moles of solute/ kg of solvent

i = 1 [Van't hoff factor]

Kf = 2.4 oC/m

dTf = -94.1 - (-93.4) = -0.7 oC

Using,

dTf -i.Kf.m

therefore

-0.7 = -1 x 2.4 x m

m = 0.292

and molality = number of moles / solvent in kg

number of moles = 0.292 mole / kg *0.050 kg

= 0.0146 moles

Molar mass = amount in g / number of moles

= 1.31g /0.0146 mole

= 89.73 g/ mole