Question

If the freezing point depression for a solution is 2. 5°c and kf = 4. 5°c/m, what is the molality of the solution?

Answer 1

Given the freezing-point depression for a solution is 2. 5 °C and the cryoscopic constant is 4.5 °c/m, the molality of the solution is 0.56 m.

What is the freezing-point depression?

Freezing-point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.

The freezing-point depression for a solution is 2. 5 °C. We can calculate the molality of the solution using the following expression.

ΔT = Kf × b

b = ΔT / Kf = 2.5 °C / (4.5 °C/m) = 0.56 m

where,

• ΔT is the freezing-point depression.
• Kf is the cryoscopic constant of the solvent.
• b is the molality of the solution.

Given the freezing-point depression for a solution is 2. 5 °C and the cryoscopic constant is 4.5 °c/m, the molality of the solution is 0.56 m.

Learn more about freezing-point depression here: brainly.com/question/14115775

Answer 2

The molality of a solution will be 0.56 m. if the freezing point depression for a solution is 2.5° C.

What is molality?

Molality is the measure of the moles of any solute in a solution per unit kg of the solvent.

Given, the freezing-point depression for a solution is 2. 5 °C

The kf is 4.5° c/m

$\rm \Delta T = Kf \times b\\\\b = \dfrac{\Delta T}{kf} \\\\b = \dfrac{2.5^ \circ C}{4.5 \circ C/m} = 0.56 m$

Where T is freezing point depression

kf is cryoscopic constant of the solvent

b is molality of the solution

Thus, the molality of the solution is 0.56 m.

Learn more about molality

brainly.com/question/20366625

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