Asolution is made by dissolving 19.5 grams of glucose (c6h12o6) in 0.245 kilograms of water. if the molal freezing point constant for water (kf) is -1.86 °c/m, what is the resulting δtf of the solution? show all he steps taken to solve this problem. will make 20

- 0.8217°C.

Explanation:

ΔTf = Kf.m,

ΔTf is the depression in the freezing point.

Kf is the molal freezing point constant for water (Kf = -1.86 °C/m).

m is the molality of glucose.

Molality (m) is the no. of moles of solute in 1.0 kg of solvent.

∴ m = (mass/molar mass) of glucose/(mass of water (kg))

∴ m = (19.5 g/180.156 g/mol)/(0.245 kg) = 0.4418 m.

∴ ΔTf = Kf.m = (-1.86 °C/m)(0.4418 m) = - 0.8217°C.

2b2t

explanation:

2b2t

fe 55.85 g/mol and h2o 18.02 g/mol.