The mass of glucose solute dissolved in the solution is 6.739 Kg.
Recall that;
ΔT = K m i
ΔT = Freezing point depression
K =Freezing point depression constant = 1.86°C/mol
m = molality of the solution
i = Van't Hoff factor = 1 (molecular solution)
We have to find the freezing point depression from;
Freezing point depression = Freezing point of pure solvent - Freezing point of solution
Freezing point of pure water = 0°C
Freezing point of solution = -5. 8 ∘C
Freezing point depression = 0°C - (-5. 8 ∘C) = 5. 8 ∘C
Now;
m = ΔT/K i
m = 5. 8 ∘C/ 1.86°C/mol × 1
m = 3.12 m
But molality = number of moles of solute/mass of solvent in Kg
Molar mass of solute = 180 g/mol
Let the mass of solute be m
3.12 = m/180/12
3.12 = m/180 × 12
m = 3.12 × 180 × 12
m = 6739 g or 6.739 Kg
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