Mass of sugar in \(5 \%\) (by mass) solution means \(5 \mathrm{~g}\) in \(100 \mathrm{~g}\) of solvent (water)
Molar mass of sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)=342 \mathrm{~g} \mathrm{~mol}^{-1}\) \(5 \%\) solution (by mass) of cane sugar in water means \(5 \mathrm{~g}\) of cane sugar in \((100-5) \mathrm{g}=95 \mathrm{~g}\) of water
No. of moles of cane sugar \(=5 / 342\)
\(\therefore \Delta T_f\) for sugar solution \(=273 \cdot 15-271=2 \cdot 15 \mathrm{~K}\)
\(\Delta T_f=K_f \times m\)
\(\Delta T_f=K_f \times 0.1537 \Rightarrow K_f=2 \cdot 15 / 0 \cdot 1537\)
\(5 \%\) glucose in water means \(5 \mathrm{~g}\) of glucose is present in 95 \(\mathrm{g}\) of water.
No. of moles of glucose \(=\frac{5}{180}\)
(Molar mass of glucose \(=180 \mathrm{~g} \mathrm{~mol}^{-1}\) )
\(\therefore\) Molality of solution \(=\frac{5 \times 1000}{180 \times 95}\)
\(=0.2926 \mathrm{~mol} \mathrm{~kg}-1\)
\(\Delta T_f=K_f \times m=\frac{2.15}{0.1537} \times 0.2926=4.09 \mathrm{~K}\)
\(\therefore\) Freezing point of glucose solution
\(=273 \cdot 15-4 \cdot 09=269 \cdot 06 \mathrm{~K}\)