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

Answer 1

Answer: (C)

$\Delta T_f=K_f\times \frac{w}{m}\times \frac{1000}{W}$
$\frac{\Delta T_{f_2}}{\Delta T_{f_1}}=\frac{m_1}{m_2}$
Here, $m_1$ (cane sugar $C_{12}{H}_{22}{O}_{11}$ ) $=342$
$m_2$ (glucose $C_6{H}_{12}{O}_6$ ) $=180$ )
$\Delta T_{f_1}=273.15-271$
$=2.15K$
$\frac{\Delta T_{f_2}}{2.15}=\frac{342}{180}$
$\Delta T_{F_2}=4.085K$
So, freezing point of glucose in water
$=273.15-4.085=269.05K$