Question

What is the percent by mass of iodine needed to reduce the freezing point of benzene to ${3.5^0}C$. The freezing point and ${K_f}$ of pure benzene are ${5.5^0}C$ and $5.12K{m^{ - 1}}$ respectively?

Answer 1

The freezing point of a solvent decreases by the addition of a solute. The change in freezing point can be calculated from molality and molal freezing point constant. The molality can be defined as the number of moles of solute in one kilogram of solvent.

Complete answer:
Freezing point is defined as the point or temperature at which the liquid starts to freeze. The freezing point of a solvent can be decreased by the addition of any solute. This phenomenon can be called depression in freezing point. The depression in freezing point can be equal to molality and molar freezing point.
Given that freezing point of pure benzene is ${5.5^0}C$
The freezing point of benzene is reduced to ${3.5^0}C$ by the addition of iodine.
Molar freezing point constant ${K_f}$ is $5.12K{m^{ - 1}}$
The change in freezing point will be obtained from subtraction of reduced freezing point from freezing point of pure benzene.
$\Delta {T_f} = 5.5 - 3.5 = {2^0}C$
But, $\Delta {T_f} = {K_f} \times m$
We know that ${K_f}$ is $5.12K{m^{ - 1}}$
M is molality will be equal to $\dfrac{2}{{5.12}} = 0.39$
This molality is equal to mass of iodine divided by molar mass of iodine
Molar mass of iodine is $254$
Thus, mass of iodine needed will be $0.39 \times 254 = 99.06gm{\left( {kg} \right)^{ - 1}}$
The volume of solution will be $1000 + 99.06 = 1099.06gm$
The mass percent of iodine is $\dfrac{{99.06}}{{1099.06}} \times 100 = 9.01\%$

Note:
The change in freezing point will be equal to molality and molar freezing point. Molality can be defined as the number of moles of solute dissolved in one kilogram of solvent. The number of moles of solute can be defined as the ratio of mass of solute to molar mass of the solute. The mass of solvent is one kilogram. Thus, molality will be equal to the number of moles of solute.