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

Step-by-step 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* 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 =
`(1.52)/(100)* 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* K_f* \frac{m_(solute)* 1000}{M_(solute)* 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* 1.86^oC/m* (2.138* 1000)/(95.2* 51.702)\\\\\text{Freezing point of solution}=-0.808^oC`

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