Answer:
4.1kg
Explanation:
I dont kwon but it was ok in the test
The mass of glucose dissolved in the solution is obtained as 4.734 Kg.
We know that;
ΔT = K m i
ΔT = Freezing point depression
K = freezing constant = 1.86 °C/m
m = molality of solution
i = Van't Hoff factor = 1 (molecular solution)
To find the freezing point depression;
Freezing point depression = Freezing point of pure solvent - Freezing point of solution
Since freezing point of water = 0°C
Freezing point depression = 0°C - (-4. 2 ∘C) = 4. 2 ∘C
The molality of the solution is obtained as follows;
m = ΔT/K i
m = 4. 2 ∘C/ 1.86 °C/m × 1
m = 2.63 m
Molality = number of moles/mass of solvent in Kg
Number of moles = mass/molar mass =
Molar mass of glucose = 180 g/mol
Let the mass of solute be m
2.63 = m/180/10
2.63 = m/1800
m = 2.63 × 1800
m = 4734 g or 4.734 Kg
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