Question

Calculate the freezing point of an aqueous solution containing $10.50g$of $MgB{r_2}$ in $200g$ of water (molar mass of $MgB{r_2} = 184g,{K_f}$ for water $= 1.86K\;kg\;mo{l^{ - 1}}$).

Answer 1

As we know that freezing point is that temperature at which the liquid and the solid forms of the solvent are in equilibrium and hence have the same vapour pressure. And the depression at freezing point is the property of decrease in freezing point when some non-volatile solute is dissolved.

Complete step-by-step solution: As we know that the depression in freezing point is one of the colligative properties that is calculated as decrease in the temperature when some non-volatile solute is added to the solution. And when liquid and solid forms of the solvent are present in equilibrium with each other and possess a similar vapour pressure that temperature is called the freezing point. And it is given by the formula as:
$\Delta {T_f} = {K_f} \times m$ where $m$ is the molality of the solute and ${K_f}$ is the molal depression constant.
Now, we are to calculate the freezing point of solution containing magnesium bromide. So let us assume that complete dissociation of magnesium bromide is taking place in the reaction. we can show this reaction as given below:
$MgB{r_2} \to M{g^{2 + }} + 2B{r^ - }$
Now, we are given that the mass of the solute is $10.50g$, molar mass of solute is $184g$and mass of solvent is $200g$ and the value of ${K_f}$ is given as $1.86K\;kg\;mo{l^{ - 1}}$. So after putting all the values in the formula of freezing point we will get:
$\Delta {T_f} = {T^o } - {T_s} = {K_f} \times \dfrac{{moles\;of\;solute}}{{weight\;of\;solvent(kg)}}$
$\Rightarrow \Delta {T_f} = {T^o } - {T_f} = 1.86 \times \dfrac{{10.50}}{{184 \times 200}} \times 1000$
$\Rightarrow \Delta {T_f} = {T^o } - {T_f} = 0.53K$
$\Rightarrow {T_f} = {T^o } - \Delta {T_f}$
$\Rightarrow {T_f} = 273 - 0.53 = 272.47K$

Therefore, the freezing point is found to be $272.47K$.

Note: Remember that the freezing point and boiling point of any substance changes when matter is dissolved in a liquid. The freezing point of that substance decreases whereas the boiling point increases. The boiling point can be calculated using the formula: $\Delta {T_b} = {K_b} \times m$where m is the molality.