Question

Constants | Periodic Table Calculate the freezing point of each of the following solutions:

Part A 0.560 mole of lactose, a nonelectrolyte, added to 1.00 kg of water Express your answer to three significant figures and include the appropriate units. nothingnothing SubmitRequest Answer Part B 44.0 g of KCl, a strong electrolyte, dissolved in 1.00 kg of water Express your answer to three significant figures and include the appropriate units. nothingnothing SubmitRequest Answer Part C 1.3 moles of K3PO4, a strong electrolyte, dissolved in 1.00 kg of water Express your answer to two significant figures and include the appropriate units.

Answer 1

A)

Lets calculate molality first

m(solvent)= 1 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.56 mol)/(1 Kg)

= 0.56 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*0.56

= 1.0416 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 1.0416

= -1.0416 oC

Answer: -1.04 oC

B)

Lets calculate molality first

Molar mass of KCl,

MM = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

mass(KCl)= 44 g

use:

number of mol of KCl,

n = mass of KCl/molar mass of KCl

=(44 g)/(74.55 g/mol)

= 0.5902 mol

m(solvent)= 1 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.5902 mol)/(1 Kg)

= 0.5902 molal

i for KCl = 2 as it dissociates into 1 K+ and 1 Cl-

lets now calculate ΔTf

ΔTf = i*Kf*m

= 2.0*1.86*0.5902

= 2.1956 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 2.1956

= -2.1956 oC

Answer: -2.20 oC

C)

Lets calculate molality first

m(solvent)= 1 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(1.3 mol)/(1 Kg)

= 1.3 molal

i for K3PO4 is 4 as it dissociates into 3 K+ and 1 PO43- ion

lets now calculate ΔTf

ΔTf = i*Kf*m

= 4.0*1.86*1.3

= 9.672 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 9.672

= -9.672 oC

Answer: -9.67 oC