What mass of glucose (C6H12O6) should be dissolved in 10. 0 kg of water to obtain a solution with a freezing point of -4. 2 ∘C?.

Answers

Answer 1

Answer:

4.1kg

Explanation:

I dont kwon but it was ok in the test

Answer 2

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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Answer:

C - to escape being eaten by the whale

I'm taking the test :)

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Answer:

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Answer:

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Explanation: That's my answer

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Answer:

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