Thank you for reporting, we will resolve it shortly
Q.
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
$\Delta T _{ f }= K _{ f } \times \frac{ w }{ m } \times \frac{1000}{ W }$
$\frac{\Delta T _{ f _{2}}}{\Delta T _{ f _{1}}}=\frac{ m _{1}}{ m _{2}}$
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
$=273.15-4.085=269.05\, K$