Calculate the freezing point of a 0.09500 m aqueous solution of glucose. the molal freezing-point-depression constant of water is is 1.86 *c/m

(a)-0.177
(b) 0.0475
(c) 0.106
(d)-0.0562
(e) -0.354

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

ΔT = T° frezzing pure solvent - T° freezing solution

(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1

T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C

i'm assuming that it's c.) red bank..