# Homework Solution: estion 1 (5 points) 17.1 g of calcium chloride is dissolved in 750, mL of a water (d =…  estion 1 (5 points) 17.1 g of calcium chloride is dissolved in 750, mL of a water (d = 1.00 g/mL) and is placed in the freezer at -2 °C overnight. Which statement best describes what would happen to the solution? O a The freezing point of the calcium chloride solution would be -1.15 C, so the solution would freeze. h The freezing point of the calcium chloride solution would be b) -0.382°C, so the solution would freeze. e) The freezing coint of the calcum chloride solution would be 1.15 c, so the solution would freeze Save 9/6/2017

1) Molar mass of CaCl2,  estion 1 (5 subject-matters) 17.1 g of calcium chloride is dissolved in 750, mL of a impart (d = 1.00 g/mL) and is placed in the benumbr at -2 °C overnight. Which assertion best describes what would fall to the repartee? O a The freezing subject-matter of the calcium chloride repartee would be -1.15 C, so the repartee would benumb. h The freezing subject-matter of the calcium chloride repartee would be b) -0.382°C, so the repartee would benumb. e) The freezing coint of the calcum chloride repartee would be 1.15 c, so the repartee would benumb Save 9/6/2017

## Expert Repartee

1)

Molar lump of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

mass(CaCl2)= 17.1 g

reckon of mol of CaCl2,

n = lump of CaCl2/molar lump of CaCl2

=(17.1 g)/(110.98 g/mol)

= 0.1541 mol

m(solvent)= 750 g

= 0.75 Kg

Molality,

m = reckon of mol / lump of solvent in Kg

=(0.1541 mol)/(0.75 Kg)

= 0.2054 molal

i coercion CaCl2 is 3

use:

Δ Tf = i*Kf*mb

Δ Tf = 3.0*1.86 oC/m * 0.2054 m

Δ Tf = 1.15 oC

So, freezing subject-matter is -1.15 oC

Since temperature is close than -1.15 oC, it earn benumb