Question

# An aqueous solution containing 34.9 g of an unknown molecular (nonelectrolyte) compound in 130.0 g of...

An aqueous solution containing 34.9 g of an unknown molecular (nonelectrolyte) compound in 130.0 g of water has a freezing point of -1.4 ∘C. Calculate the molar mass of the unknown compound.

#### Homework Answers

Answer #1

Freezing point depression is

∆Tf=Kf x m

Where ∆Tf=Tf(pure solvent) - Tf(solution)

Kf=freezing point constant,

m=molality=(mass/molar mass)(1000/mass of solvent)

Given mass=34.9 g and mass of solvent (water)=130 g,

Molar mass of nonelectrolyte=?,

Tf(solution)=-1.4°C, Tf(solvent)=0°C, Kf of water=1.86°C/m=1.86°C Kg/mol.

Therefore

(0°C-(-1.4°C))=(1.86°C Kg/mol)(34.9g/molar mass) (1000/130)

Molar mass=(1.86 x 34.9x 1000)/(1.4x130)=356.67 g/mol

Therefore molar mass of non-electrolyte=356.67 g/mol.

Please let me know if you have any doubt. Thanks.

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