Question

How do you do freezing point depression problems $?$

Answer 1

Hint : Freezing-point depression is the decrease of the freezing point of a solvent on the addition of a non-volatile solute and is given by the equation $\Delta {T_f} = i \times {k_f} \times m$ where $i$ is the Van’t Hoff Factor, ${k_f}$ is the freezing point depression constant and $m$ represents the molality of the solution.

Complete Step By Step Answer:
Freezing point depression problems are generally given for the calculation of the number of moles of non-volatile solute added to the solution so that there is a depression in the freezing point.
We know that the depression of freezing point is given by the equation $\Delta {T_f} = i \times {k_f} \times m........\left( 1 \right)$ where $i$ is the Van’t Hoff Factor, ${k_f}$ is the freezing point depression constant and $m$ represents the molality of the solution.
Now, the solution must have a solute and a solvent.
Calculate the Van’t Hoff Factor $\left( i \right)$ from the dissociation or association of the solute particles.
${k_f}$ or the freezing point depression constant can be known from literature.
Now since $m$ represents the molality of the solution it can be given by
$m = \dfrac{{No.\,of\,moles\,solute}}{{Mass\,of\,solvent\,\left( {in\,kg} \right)}}$ .
Mass of the solvent must be given in the question and convert it to proper units.
$\Delta {T_f}$ can either be mentioned in the question or calculate it by subtracting the final temperature from initial temperature.
Put all the values in equation $\left( 1 \right)$ and calculate the number of moles of solute present in the solution.
Now, multiply the molecular weight of solute with the number of moles of solute in solution to get the amount of solute added for depression in freezing point.

Note :
For these types of questions you should take proper care of the units. Do not mix up the units and make sure to do the conversions wherever necessary. Try to proceed with the calculations in a stepwise manner as much as possible as this reduces the chances of doing any error.