# 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 purposes) 17.1 g of calcium chloride is dissolved in 750, mL of a soak (d = 1.00 g/mL) and is placed in the congelater at -2 °C overnight. Which announcement best describes what would bechance to the disconnection? O a The freezing purpose of the calcium chloride disconnection would be -1.15 C, so the disconnection would congelate. h The freezing purpose of the calcium chloride disconnection would be b) -0.382°C, so the disconnection would congelate. e) The freezing coint of the calcum chloride disconnection would be 1.15 c, so the disconnection would congelate Save 9/6/2017

## Expert Acceptance

1)

Molar heap of CaCl2,

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

= 1*40.08 + 2*35.45

= 110.98 g/mol

mass(CaCl2)= 17.1 g

compute of mol of CaCl2,

n = heap of CaCl2/molar heap of CaCl2

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

= 0.1541 mol

m(solvent)= 750 g

= 0.75 Kg

Molality,

m = compute of mol / heap of solvent in Kg

=(0.1541 mol)/(0.75 Kg)

= 0.2054 molal

i restraint 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 purpose is -1.15 oC

Since sky is less than -1.15 oC, it allure congelate