Question

A $10$% solution (by mass) of sucrose in water has a freezing point of $269.15K$. Calculate the freezing point of $10$% glucose in water if the freezing point of pure water is $273.15K$.
(Given : molar mass of sucrose $=342gmo{l}^{-}$, Molar mass of glucose $=180gmo{l}^{-1}$)

• A. 275. 53 K
• B. 265. 53 K
• C. 271. 32 K
• D. 282.43 K

Answer 1

The correct option is B 265. 53 K

For sucrose,

$\Delta T_f=K_{f}m$

$\Delta T_f=K_f\times \frac{10}{342}\times \frac{1}{Massofsolvent}⟶\left(1\right)$

For glucose,

$\Delta T_f=K_f\times \frac{10}{180}\times \frac{1}{Massofsolvent}⟶\left(2\right)$

Both have the same mass of solvent

Divide equation (1) & (2)

$\Rightarrow \frac{\Delta T_f\left(sucrose\right)}{\Delta T_f\left(glucose\right)}=\frac{180}{342}$

$\Rightarrow \frac{T_f^o-T_f\left(sucrose\right)}{T_f^o-T_f\left(glucose\right)}=\frac{180}{342}$

$\Rightarrow \frac{273.15-269.15}{273.15-T_f}=\frac{180}{342}$

$\Rightarrow \frac{4}{273.15-T_f}=0.526$

$\Rightarrow \frac{4}{0.526}=273.15-T_f$

$\Rightarrow T_f=273.15-\frac{4}{0.526}$

$\Rightarrow T_f=273.15-7.6$

$\Rightarrow T_f=265.54K$