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PLEASE ANSWER ALL PARTS AND SHOW THE WORK CLEARLY. 2. A solution containing 2.00

ID: 1082150 • Letter: P

Question

PLEASE ANSWER ALL PARTS AND SHOW THE WORK CLEARLY.

2. A solution containing 2.00 g of an unknown substance in 50.0 g of cyclohexane freezes at 1.05°C. The normal freezing point of cyclohexane is 6.60°C and Kf = 20.4 °C/m. Both substances are molecular.

a. What is the value of i? __________

b. Calculate the molality of the solution using T = Kf • m • i. Show all work.

c. The units of molality are moles/kg. Moles of what over kilograms of what? Be more specific than solute and solvent – label the compounds!

Moles of ________ over kilograms of _______

d. How many moles of the unknown substance are in this solution? Show all work.

e. What is the molar mass of the unknown substance? Show all work

Explanation / Answer

a) since cyclohexane is a non-elctrolyte

i = 1

b) we know that

dT = freezing point of cyclohexane - freezing point of solution

dT = 6.6 - 1.05

dT = 5.55

now

dT = Kf x m x i

5.55 = 20.4 x m x 1

m = 0.272

so

the molality is 0.272 mol/kg

c)

moles of the given unknown substance over kilograms of cyclohexane

d)

given cyclohexane = 50 g = 0.05 kg

now

0.272 = moles of unknown substance / 0.05 kg

moles of unknown substance = 0.0136 mol

e)

now

molar mass = mass / moles

molar mass of the unknown substance = 2 g / 0.0136 mol

molar mass of the unknown substance = 147.02 g/mol

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