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

Explanation

`5%` solution by mass means 5 g of solute is present in 100 g of solution
`therefore `Mass of solvent (Water) = 95 g
`"Molality of sugar solution "=(5)/(342)xx(1000)/(95)=0.154" "("mol. Mass of sugar "C_(12)H_(22)O_(11)=342)`
`DeltaT_(f)" for sugar solution "=273.15-271=2.15^(@), DeltaT_(f)=K_(f)xxm therefore K_(f)=(2.15)/(0.154)`
`"Molality of glucose solution "=(5)/(180)xx(1000)/(95)=0.292" "("mol. mass of glucose "C_(6)H_(12)O_(6)=180)`
`therefore" "DeltaT_(f)" (Glucose)" =K_(f)xxm=(2.15)/(0.154)xx0.292=4.08`
`therefore" Freezing point of glucose solution "=273.15-4.08="269.07 K."`