PLEASE ANSWER ALL PARTS AND SHOW THE WORK CLEARLY. 2. A solution containing 2.00
ID: 578554 • 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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