Question

A $5$% solution (by mass) of cane sugar in water has freezing point of $271K$ and freezing point of pure water is $273.15K$. The freezing point of a $5$% solution (by mass) of glucose in water is :

• A. $271K$
• B. $273.15K$
• C. $269.07K$
• D. $277.23K$

Answer 1

Answer: (C)

$269.07K$

From the relation $△T_f=K_f\times \frac{w}{m}\times \frac{1000}{W}$

It is obvious that $△T_f\propto \frac{1}{m}$

$\therefore \frac{△T_{f_1}}{△T_{f_2}}=\frac{m_2}{m_1}$

Cane sugar solution Glucose solution
$△T_{f_1}=273.15=271$ $△T_{f_2}=?$
$m_1=342$ $M_2=180$

Hence, $\frac{2.15}{△T_{f_2}}=\frac{180}{342}$

$△T_{f_2}=\frac{342\times 2.15}{180}=4.085K$

So, freezing point of glucose solution $=273.15-4.085$
$=269.05K$