Question

Given: A 50.0-mL solution is initially 1.49% MgCl2 by mass and has a density of 1.05 g/mL.

What is the freezing point of the solution after you add an additional 1.36 g MgCl2? (Use i = 2.5 for MgCl2.)

Express your answer using two significant figures. Answer: T= (blank) degrees celsius

Answer 1

Volume of given solution = 50.0 mL

density of solution = 1.05 g/mL

mass of the given solution grams = 52.5 grams

mass of in the given solution = (1.49% by mass)*(52.5grams of solution)=0.78225 grams

mass of water in the given solution = 52.5 grams of solution - 0.78225 grams of =51.71775 grams

after adding additional 1.36 grams of , total mass is = 0.78225+1.36 = 2.14225 grams

now molar mass of is 95.21 grams

therefore number of moles of = 2.14225/95.21=0.0225 moles of

mass of water = 51.71775 grams = 0.05171 Kg

therefore molality of = (0.0225 moles of )/(0.0517 Kg of water) = 0.4352 molal

now change in temperature, dT is given by,

i = 2.5 for

= molal freezing point depression constant = 1.86 C/molal

therefore there is 2.024 C drop in temperature

New freezing point = T = 2.024 degree celsius