Question

If 33.9g NaCl are mixed into water and the total mass is 578g, what is the CHANGE in freezing if Kb= - 1.82C/M (molal)? Assume NaCl does not dissociate in solution.

Answer 1

-1.82 °C

Step-by-step explanation:

Step 1: Given data

• Mass of NaCl (solute): 33.9 g

• Mass of water (solvent): 578 g = 0.578 kg

• Freezing point depression constant for water (Kb): -1.82 °C/m

Step 2: Calculate the molality of the solution

We will use the following expression.

m = mass of solute / molar mass of solute × kg of solvent

m = 33.9 g / 58.44 g/mol × 0.578 kg

m = 1.00 m

Step 3: Calculate the freezing point depression (ΔT)

The freezing point depression is a colligative property that, for a non-dissociated solute, can be calculated using the following expression:

ΔT = Kb × m

ΔT = -1.82 °C/m × 1.00 m

ΔT = -1.82 °C