Question

An aqueous solution containing 36.5 g  of an unknown molecular (non-electrolyte) compound in 151.5 g of water was found to have a freezing point of -1.5 ∘C.

Answer 1

The formula for the freezing point is given by,

T=K_f x m

where T is the freezing temp i.e -1.5-degree Celsius

K_f for water is 1.86-degree Celsius / Molal

m is the molality

so, putting the values

1.5 = m x k_f

=m x 1.86

m=1.5/1.86

m = 0.806

molality=moles of solute/kg of solvent------(1)

solvent here is water so, 151.5g = 0.1515Kg

putting values in (1)

0.806=molesof solute/0.1515Kg

moles of solute=0.1221 moles

no.of moles = given mass/molar mass

molar mass=given mass/no.of moles

Molar mass = 36.5 / 0.1221 = 298.9g/mol