Answer:
3.9°C/m is the molal freezing-point constant of the unknown solvent.
Explanation:
[tex]\Delta T_f=i\timesK_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
i = van't Hoff factor
[tex]K_f[/tex] = freezing point constant
m = molality
we have :
[tex]K_f[/tex] =?
[tex]m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}[/tex]
[tex]m=\frac{2 mol}{1 kg}=2 m[/tex]
i = 1 ( non electrolyte)
[tex]\Delta T_f=7.8^oC[/tex]
[tex]7.8^oC=1\times K_f\times 2 m[/tex]
[tex]K_f=3.9 ^oC/m[/tex]
3.9°C/m is the molal freezing-point constant of the unknown solvent.
The solution has a freezing constant of 3.9 °C/m. The solvent is most likely acetic acid.
Using the relation;
ΔT = K m i
ΔT = 7.8°C
m = number of moles of solute/ Mass of solution 2 moles/1 Kg = 2 m
i = Non - electrolyte solution = 1
7.8°C = K. 2 m. 1
K = 7.8°C/2 m. 1
K = 3.9 °C/m
The solvent is most likely acetic acid.
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