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

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.