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

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).

The freezing point of solution is -0.808°C

Explanation:

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

Density of solution = 1.05 g/mL

Volume of solution = 50 mL

Putting values in above equation, we get:

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 =

The equation used to calculate depression in freezing point follows:

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

Or,

where,

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

i = Vant hoff factor = 2.5

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

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

= Molar mass of solute (magnesium chloride) = 95.2  g/mol

= Mass of solvent (water) = [52.5 - 0.798] g = 51.702 g

Putting values in above equation, we get:

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

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

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