Question

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

The Correct Option is 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 $C _{6} H _{12} O _{6}$ ) $=180$ ) $\Delta T _{ f _{1}}=273.15-271$ $=2.15 \,K$ $\frac{\Delta T _{ f _{2}}}{2.15}=\frac{342}{180}$ $\Delta T _{ F _{2}}=4.085 \,K$ So, freezing point of glucose in water $=273.15-4.085=269.05\, K$