1.8 ~g of glucose (molar mass 180 ~g ~mol ⁻¹ ) is dissolved in 0.1 ~kg of water. The freezing point of the solution ( in ^ C ) is (K_f . for water .=1.86 ~K ~kg ~mol ⁻¹ )

Step-by-step solution

Depression in freezing point is given as T=i K_f m where, i= van't Hoff factor gathered K_f= molal freezing point depression constant gathered m= molality Molar mass of glucose ( C ₆ H ₁₂ O ₆ )=12 6+12 1+6 16=180 ~g / mol = 1.8 180 1 0.1 =0.1 ~m T= I 1.86 ~K ~kg / mol 0.1 ~m =0.186^ C Freezing point of solution = Freezing point of water - Depression in freezing point =0^ C -0.186^ C =-0.186^ C