Will give ! a solution is made by dissolving 21.5 grams of glucose (c6h12o6) in 255 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.

Hey there!:

ΔTf = Kf * m

Molar mass glucose = 180 g/mol

number of moles glucose:

n = mass of solute / molar mass

n = 21.5 / 180

n = 0.119 moles glucose

Amount of solvent in kg = 255/1000 = 0.255 Kg

Molality = number of moles / solvent

m = 0.119 / 0.255

m = 0.466 moles/kg

Kf for water = - 1.86 ºC/*m

Therefore:

ΔTf = Kf * m

ΔTf = (-1,86) * 0.466

ΔTf = -0.86676 ºC

hope this helps!

ΔTf = Kf * m Molar mass glucose = 180 g/molthe number of moles glucose:n = mass of solute / molar massn = 21.5 / 180n = 0.119 moles glucoseAmount of solvent in kg = 255/1000 = 0.255 Kg Molality = number of moles / solventm = 0.119 / 0.255 m = 0.466 moles/kgKf for water = - 1.86 ºC/*mTherefore:ΔTf = Kf * m ΔTf = (-1,86) * 0.466ΔTf = -0.86676 ºC

this is to the best of my brain power and i hope this helps with your problem

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml