Question

A solution of sucrose (molar mass = ${\text{342 g mo}}{{\text{l}}^{{\text{ - 1}}}}$) has been prepared by dissolving ${\text{68}}{\text{.5 g}}$ of sucrose in 1000 g of water. The freezing point of the solution obtained will be:
[${K_f}$ for water = ${\text{1}}{\text{.86 K kg mo}}{{\text{l}}^{{\text{ - 1}}}}$]
A. $ - 0.372{{\text{ }}^{\text{o}}}{\text{C}}$
B.$0.520{{\text{ }}^{\text{o}}}{\text{C}}$
C. ${\text{ + 0}}{\text{.372}}{{\text{ }}^{\text{o}}}{\text{C}}$
D. ${\text{0}}{\text{.570}}{{\text{ }}^{\text{o}}}{\text{C}}$

Answer 1

Hint: To solve this question, we should have knowledge about the effect of impurities on the physical properties of a solution. The change in freezing point due to addition of impurities in a solution is a colligative property and can be calculated from the depression in freezing point formula (given below). Substituting the given values in this equation, we can calculate the change in freezing point of the solution. From this, we can calculate the new freezing point of the solution.

Formula used: $\Delta {T_f} = {K_f} \times \dfrac{{w \times 1000}}{{{M_0} \times W}}$ (Eq. 1)
where $\Delta {T_f}$ is the change in freezing point of the solution, ${K_f}$ is freezing constant (given), $w$ is weight of the solute (in g), ${M_0}$ is the molar mass of the solute (in ${\text{g mo}}{{\text{l}}^{{\text{ - 1}}}}$ ), $W$ is the weight of the solution (in g).


Complete step by step answer:
Colligative properties are the properties which depend only on the number of solute particles present in the solution and not on their nature. The main colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. In the question, we have to find a new freezing point of the solution. For that, we shall first find the depression in the freezing point. So, by substituting all the given values in Eq. 1, we get:
$\Delta {T_f} = 1.86 \times \dfrac{{68.5 \times 1000}}{{342 \times 1000}}$
Solving this for $\Delta {T_f}$ we get,
$\Delta {T_f} = 1.86 \times 0.2$
$ \Rightarrow \Delta {T_f} = {0.372^o}C$
Now, as we know that the freezing point of water is ${\text{0}}{{\text{ }}^{\text{o}}}{\text{C}}$ , we can calculate the new freezing point of the solution by:
$\Delta {T_f} = {T_{final}} - {T_{initial}}$
So, the final temperature of the solution ( ${T_{final}}$ ) will be:
${T_{final}} = \Delta {T_f} + {T_{initial}}$
$ \Rightarrow {T_{final}} = 0.372 + 0 = {0.372^o}C$

$\therefore $ The freezing point of the solution will be $ + {0.372^o}C$ , i.e. Option C .

Note:
The Eq. 1 can also we written as $\Delta {T_f} = {K_f} \times {\text{molality}}$, so if molality of a solution is given, we can directly replace that value in this formula to calculate the depression in freezing point. In cases where the impurity can dissociate into ions, the formula is modified into $\Delta {T_f} = i \times {K_f} \times {\text{molality}}$ where $i$ is the number of ions formed in the solution after dissociation. For example, in case of ${\text{NaCl}} \to {\text{N}}{{\text{a}}^{\text{ + }}}{\text{ + C}}{{\text{l}}^{\text{ - }}}$, $i = 2$ as 2 ions are formed in the solution.