A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a solution made of of alanine () dissolved in of .

The question is incomplete, the complete question is;

A certain liquid X has a normal freezing point of 5.40 °C and a freezing point depression constant K=3.05 ?kg mol solution made of 46.1g of alanine (C3H,NO2) dissolved in 750. g of i. Round your answer to 3 significant digits.

3.3 °C

Explanation:

If ∆T= K m I

Where;

∆T = freezing point depression

K = freezing point constant

m= morality of solution

i= Van't Hoff factor

Number of moles of solute = 46.1 g/89.09 g/mol = 0.52 moles

Molality of solution = 0.52 moles /750 ×10-3Kg

= 0.69 m

∆T= 3.05 × 0.69 × 1

∆T= 2.10 °C

Freezing point of solution the solution = freezing point of pure solvent - freezing point depression

= 5.40°C - 2.10 °C

= 3.3 °C