Respuesta :
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:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.05 g/mL
Volume of solution = 50 mL
Putting values in above equation, we get:
[tex]1.05g/mL=\frac{\text{Mass of solution}}{50mL}\\\\\text{Mass of solution}=(1.05g/mL\times 50mL)=52.5g[/tex]
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 = [tex]\frac{1.52}{100}\times 52.5=0.798g[/tex]
The equation used to calculate depression in freezing point follows:
[tex]\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}[/tex]
To calculate the depression in freezing point, we use the equation:
[tex]\Delta T_f=iK_fm[/tex]
Or,
[tex]\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)}}[/tex]
where,
Freezing point of pure solution (water) = 0°C
i = Vant hoff factor = 2.5
[tex]K_f[/tex] = molal freezing point elevation constant = 1.86°C/m
[tex]m_{solute}[/tex] = Given mass of solute (magnesium chloride) = [0.798 + 1.34] g = 2.138 g
[tex]M_{solute}[/tex] = Molar mass of solute (magnesium chloride) = 95.2 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = [52.5 - 0.798] g = 51.702 g
Putting values in above equation, we get:
[tex]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[/tex]
Hence, the freezing point of solution is -0.808°C
