Asolution is made by dissolving 10.20 grams of glucose (c6h12o6) in 355 grams of water. what is the freezing point depression of the solvent if the freezing point constant is -1.86 °c/m? show all of the work needed to solve this problem.

0.297 °C

Step-by-step explanation:

The formula for the freezing point depression ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

  Mass of water = 355 g

                 ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

(a) Moles of glucose

n = 10.20 g × (1 mol/180.16 g)

  = 0.056 62 mol

(b) Kilograms of water

m = 355 g × (1 kg/1000 g)

   = 0.355 kg

(c) Molal concentration

b = moles of solute/kilograms of solvent

  = 0.056 62 mol/0.355 kg

  = 0.1595 mol·kg⁻¹

(d) Freezing point depression

ΔT_f = 1 × 1.86 × 0.1595

        = 0.297 °C

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2