Respuesta :
The relation between freezing point depression of a solution and molality is,
Δ [tex] T_{f} [/tex] = i [tex] K_{f} m [/tex]
i = Van't Hoff factor = 3 as MgCl2 gives 3 ions ([tex] MgCl_{2}(aq)---> Mg^{2+} (aq) + 2 Cl^{-}(aq) [/tex])
[tex] K_{f} [/tex] is the freezing point molar constant of water = [tex] 1.86 ^{0}C/m [/tex]
m is the molality of solution
Calculating the molality of solution:
[tex] 7.15 g MgCl_{2}*\frac{1 mol MgCl_{2}}{95.21 g MgCl_{2}}*\frac{1}{100 g}*\frac{1000 g}{1kg} [/tex]
= 0.751 m
Plugging in the values of m and [tex] K_{f} [/tex] to get freezing point of the solution:
Δ[tex] T_{f}= 3 (1.86^{0}C/m)(0.751 m) [/tex]
[tex] T_{f} (water) - T_{f}(solution) = 4.19^{0}C [/tex]
[tex] 0^{0}C - T_{f}(solution) = 4.19^{0}C [/tex]
[tex] T_{f}(solution) = - 4.19^{0}C [/tex]
The freezing point is the temperature at which the solution gets frozen. The freezing point of the solution is -4.19 °C.
What is freezing point depression?
Freezing point depression is the property of the solution that is given by the molality, cryoscopic constant, and the van 't Hoff factor. The formula is given as:
Δ T (freezing)= i × Kf × m
The chemical reaction can be shown as MgCl₂ → Mg²⁺ 2Cl⁻
The van 't Hoff factor will be 3 as three ions are produced in the above reaction.
The molality (m) of the solution is calculated as:
(7.15 × 1000) ÷ (95.21 × 100) = 0.751 m
Substituting values in the equation the freezing point is calculated as:
Δ T (freezing)= 3 (1.86)(0.751)
0 - T (solution) = 4.19
T (freezing) (solution) = -4.19 °C
Therefore, -4.19 °C is the freezing point of the solution.
Learn more about freezing point, here:
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