Question

What is the freezing point of the solution after you add an additional 1.34 g (Use i = 2.5 for MgCl2)

Answer 1

The question is incomplete, here is the complete question:

A 50 mL solution is initially 1.52% 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 1.37 g MgCl₂? (Use i = 2.5 for MgCl₂).

Answer: The freezing point of solution is -0.808°C

Explanation:

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.05 g/mL

Volume of solution = 50 mL

Putting values in above equation, we get:

$1.05g/mL=\frac{\text{Mass of solution}}{50mL}\\\\\text{Mass of solution}=(1.05g/mL\times 50mL)=52.5g$

We are given:

Percentage of magnesium chloride in the solution = 1.52 %

Mass of magnesium chloride in the solution = 1.52 % of 52.5 g = $\frac{1.52}{100}\times 52.5=0.798g$

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

i = Vant hoff factor = 2.5

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (magnesium chloride) = [0.798 + 1.34] g = 2.138 g

$M_{solute}$ = Molar mass of solute (magnesium chloride) = 95.2  g/mol

$W_{solvent}$ = Mass of solvent (water) = [52.5 - 0.798] g = 51.702 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{2.138\times 1000}{95.2\times 51.702}\\\\\text{Freezing point of solution}=-0.808^oC$

Hence, the freezing point of solution is -0.808°C