Freezing point of solution =271K

5% solution by mass so take total mass of solution = 100 gram

Then 5%  of 100 gram of glucose will  = 5 g

The formula of deviation from freezing point is

∆T_f = K_f × m  ……(1)

Change in freezing point

∆T_f   =  freezing point of water – freezing point of solution

=273.15  - 271  

= 2.15

Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol

Molar mass of cane sugar (C₁₂H₂₂O₁₁) =342

Number of moles  of (C6H12O6)=  5/180  = 0.028 moles

Number of moles of (C12H22O11)= 5/ 342 = 0.0146 moles

Mass of solvent = total mass – mass of solute

                         = 100 – 5 = 95 g

To convert in Kg divide by 1000 we get mass of solvent  = 95/1000 = 0.095g

Molality of  cane sugar = 0.0146/0.095 = 0.154 m

Molality of  C6H12O6 (m)  = 0.028/0.095  =0.29 m

Plug these value in equation (1) we get

Plug the values of cane sugar we get  

2.15= K_f ×0.154 ……………..(1)

Divide by 0.154 we get

K_f = 13.97

Plug the value of glucose we get

∆T_f = K_f×m

∆T_f = 13.97× 0.29

∆T_f = 4.08

Now new freezing point  = freezing point of water  - ∆T_f

                                          =273.15 -  4.08

                                          =269.07 C