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

Hint: We know the boiling point and the freezing point of water are 100° C and 0° C respectively. Therefore, $\mathrm{\Delta }{T}_b+\mathrm{\Delta }{T}_f={105}^{\circ }-{100}^{\circ }=5^{\circ }$ (where $\mathrm{\Delta }{T}_b$ is the elevation in boiling point and $\mathrm{\Delta }{T}_f$ is the depression in freezing point).
Again, we know that $\mathrm{\Delta }{T}_f=k_f\times m$ and $\mathrm{\Delta }{T}_b=k_b\times m$ 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: $\mathrm{\Delta }{T}_f=k_f\times m$
$\mathrm{\Delta }{T}_b=k_b\times m$
Where,
$\mathrm{\Delta }{T}_b$ is the elevation in boiling point,
$\mathrm{\Delta }{T}_f$ 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, $\mathrm{\Delta }{T}_f=k_f\times m$ and $\mathrm{\Delta }{T}_b=k_b\times m$, Where,
$\mathrm{\Delta }{T}_b$ is the elevation in boiling point,
$\mathrm{\Delta }{T}_f$ 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, $\mathrm{\Delta }{T}_b+\mathrm{\Delta }{T}_f=\left(k_b+k_f\right)\times m$
$\Rightarrow \mathrm{\Delta }{T}_b+\mathrm{\Delta }{T}_f=\left(0.51+1.86\right)\times m$
$\Rightarrow \mathrm{\Delta }{T}_b+\mathrm{\Delta }{T}_f=2.37\times m$
Now, we know that the boiling point and the freezing point of water are 100° C and 0° C respectively.
Therefore, $\mathrm{\Delta }{T}_b+\mathrm{\Delta }{T}_f={105}^{\circ }-{100}^{\circ }=5^{\circ }$
$\Rightarrow 5=2.37\times m$
$\Rightarrow m={\displaystyle \frac{5}{2.37}}=2.11$
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 $={\displaystyle \frac{2.11}{1000}}\times 100=0.211$ 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.