Question

Calculate the freezing point and boiling point of each of the following solutions: the freezing point of the solution: 174 g of sucrose, C₁₂H₂₂O₁₁, a nonelectrolyte, dissolved in 1.35 kg of water (K_f=1.86∘C) Express your answer using one decimal place.

Answer 1

Lets calculate molality first

Molar mass of C12H22O11 = 12*MM(C) + 22*MM(H) + 11*MM(O)

= 12*12.01 + 22*1.008 + 11*16.0

= 342.296 g/mol

mass(C12H22O11)= 174 g

we have below equation to be used:

number of mol of C12H22O11,

n = mass of C12H22O11/molar mass of C12H22O11

=(174.0 g)/(342.296 g/mol)

= 0.5083 mol

m(solvent)= 1.35 Kg

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(0.5083 mol)/(1.35 Kg)

= 0.3765 molal

lets now calculate deltaTf

deltaTf = Kf*m

= 1.86*0.3765

= 0.7004 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 0.7004

= -0.700 oC

Answer: -0.7 oC