Solution:
Part C)
Cm = 0.0026 m
ΔTf = 0.0145 °C
1 ppm = 1 mg / L
Thus,
250 ppm = 250 mg / L = 0.250 g / L
Since,
Molality (Cm) = Mass /Molar mass x Mass of water in kg
(Since, density of water = 1 g/mL
Hence, volume of solution = mass of solution
Mass of solution = mass of water + mass of MgCl2
Therefore,
Mass of water (solvent) = 1000 g - 0.250 g = 999.75 g )
Molality = 0.250 g / 95.21 g mol-1 x 0.999 g
C m = 0.0026 m
Since,
ΔTf = i Kf Cm
i for MgCl2 = 3 (due to one Mg2+ and two Cl- ions)
.ΔTf = 3 x 1.86 °C/m x 0.0026 m
= 0.0145 °C
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