Question

Part E - How many grams of KCl must be added to depress the freezing point of 1.00 kg of water to a temperature of -1.4 C? Show your work.

Answer 1

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

we have below equation to be used:

delta Tf = i*Kf*mb

1.4 = 2.0*1.86 *mb

mb= 0.3763 molal

mass of solvent = 1.00 Kg

we have below equation to be used:

number of mol,

n = Molality * mass of solvent in Kg

= (0.3763 mol/Kg)*(1 Kg)

= 0.3763 mol

Molar mass of KCl = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

we have below equation to be used:

mass of KCl,

m = number of mol * molar mass

= 0.3763 mol * 74.55 g/mol

= 28.1 g

Answer: 28.1 g