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.
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,
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.
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The molality of a solution will be 0.56 m. if the freezing point depression for a solution is 2.5° C.
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
[tex]\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[/tex]
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.
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