Question

What mass of glucose (C6H12O6) should be dissolved in 12. 0 kg of water to obtain a solution with a freezing point of -5. 8 ∘C?.

Answer 1

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

Learn more: brainly.com/question/6249935

Answer 2

Answer:

6.7 kg

Explanation:

(0) - (-5.8)= 5.8ºC

5.8 / 1.86 = 3.12 m C6 H12 O6

The molar mass of C6 H12 O6 is: 180.156 g/mol

3.12 * 12 kg * 180.156 g/mol = 6,745.041 g = 6.7 kg.