asap help please in chemistry

Question

What is the freezing point of a solution made from 200 g of glucose (C₆H₁₂O₆ ) and 350 g of benzene? The normal freezing point of benzene is 5.5 °C, and K_f, benzene = 5.12 ° C/m.



Answer 1

The answer is -10.75 ⁰C

200 grams of glucose is 1.11 moles (200g/180 g, 180 g/mo is the molar mass of the compound)

Molality of the solution is 1.11/0.350 kg (convert grams of Benzene into kilograms to find the molality), the molality of the solution is 3.17.

Change in the freezing point= K_f X molality =5.12 X 3.17 = 16.25 ⁰C

Freezing point of the solution= Freezing point of the solvent - depression in the freezing point = 5.5 ⁰C - 16.25 ⁰C = 10.75 ⁰C