Question

How many grams of sucrose (M. wt = 342) should be dissolved in 100 g water in order to produce a solution with a 105°C difference between freezing point and boiling point temperature? (${K_b}$= 0.51 and ${K_f}$= 1.86)
A. 34.2 g
B. 72 g
C. 342 g
D. 460 g

Answer 1

We know the boiling point and the freezing point of water are 100° C and 0° C respectively. Therefore, $\Delta {T_b} + \Delta {T_f} = {105^ \circ } - {100^ \circ } = {5^ \circ }$ (where $\Delta {T_b}$ is the elevation in boiling point and $\Delta {T_f}$ is the depression in freezing point).
Again, we know that $\Delta {T_f} = {k_f} \times m$ and $\Delta {T_b} = {k_b} \times m$ where m is the molality of the solution. Find m.
The molecular weight of sucrose is given, the molality (m) is known now. Hence find the required mass of sucrose.

Formula used: $\Delta {T_f} = {k_f} \times m$
$\Delta {T_b} = {k_b} \times m$
Where,
$\Delta {T_b}$ is the elevation in boiling point,
$\Delta {T_f}$ is the depression in freezing point,
${K_b}$= 0.512°C.kg/mol is the boiling point elevation constant,
${K_f}$=1.86°C.kg/mol is the freezing point depression constant, and
m = molality of the solution.

Complete step by step answer:
Let the molality of the solution be m.
Now, $\Delta {T_f} = {k_f} \times m$ and $\Delta {T_b} = {k_b} \times m$, Where,
$\Delta {T_b}$ is the elevation in boiling point,
$\Delta {T_f}$ is the depression in freezing point,
${K_b}$= 0.512°C.kg/mol is the boiling point elevation constant,
${K_f}$=1.86°C.kg/mol is the freezing point depression constant, and
m = molality of the solution.
Therefore, $\Delta {T_b} + \Delta {T_f} = \left( {{k_b} + {k_f}} \right) \times m$
$\Rightarrow \Delta {T_b} + \Delta {T_f} = \left( {0.51 + 1.86} \right) \times m$
$\Rightarrow \Delta {T_b} + \Delta {T_f} = 2.37 \times m$
Now, we know that the boiling point and the freezing point of water are 100° C and 0° C respectively.
Therefore, $\Delta {T_b} + \Delta {T_f} = {105^ \circ } - {100^ \circ } = {5^ \circ }$
$\Rightarrow 5 = 2.37 \times m$
$\Rightarrow m = \dfrac{5}{{2.37}} = 2.11$
Now, molality of a solution is given by the number of moles of solute present in 1 kg of solvent.
Hence, a 2.11 molal solution means 2.11 moles of solute is present in 1000 gm of solvent.
Here the mass of water is 100 gm.
So, 100 gm of solvent (water) of a 2.11 (m) solution will contain $= \dfrac{{2.11}}{{1000}} \times 100 = 0.211$ moles of solute (sucrose).
Again, the gram molecular weight of sucrose is given as 342 gm.
Required mass of 0.211 moles of sucrose is $= 342 \times 0.211 = 72.2{\text{ gm}}$

Hence, the correct option is (B).

Note: Note that, the molality of a solution is given by the number of moles of solute present in 1 kg of solvent. Here, the mass of solvent is 100 gm. Therefore we need to find the number of moles of sucrose present in 100 grams of solvent.