A 5% acqueous solution by mass of sucrose has a freezing point of 272.86K. find the molal dippression constant of water?

Question

Answer

Explanation:

Definition: Molal depression constant is defined as the decrease in the freezing point of a solvent by one molal solution of a non-volatile solute.

Formula: ΔTf = Kf x m

Where,
ΔTf = depression in freezing point
Kf = molal depression constant
m = molality of the solution

Given:
Mass of sucrose = 5% (by mass)
Freezing point depression = 272.86 K

Calculations:
We first need to calculate the molality of the solution.

Mass of sucrose = 5% of solution
Mass of solvent = 100% - 5% = 95% of solution

Let's assume we have 100 g of solution. Then,
Mass of sucrose = 5 g
Mass of solvent (water) = 95 g

Molar mass of sucrose (C12H22O11) = 342.3 g/mol
Number of moles of sucrose = mass/molar mass = 5/342.3 = 0.0146 mol

Molality (m) = moles of solute/mass of solvent (in kg)
Mass of solvent (in kg) = 95/1000 = 0.095 kg
Molality (m) = 0.0146/0.095 = 0.1537 mol/kg

Now, we can use the given freezing point depression and molality to calculate the molal depression constant.

Kf = ΔTf/m
Kf = 272.86/0.1537
Kf = 1775.96 K kg/mol

Answer:
The molal depression constant of water is 1775.96 K kg/mol.

Top Courses for Class 12

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams