Please show full work and reasoning, thanks! Colligative Properties - Freezing a

Question

Please show full work and reasoning, thanks!

Colligative Properties - Freezing and Boiling Points of Solutions Calculate the freezing point of a solution of 3.95 g of naphthalene (C10Hs) in 12.20 g of benzene. The freezing point of pure benzene is 5.5°C. Kf= 5.12°C kg/mol. oc 1pts Submit Answer Tries o/5 Calculate the boilina point of the above solution. The boiling point of pure benzene is 80.1°C. Kb = 2.53°C kg/mol 1pts Submit Answer Tries 0/5 cf Zumdahl "Chemical Principles" 8th ed., sec17.5 "Boiling-Point Elevation and Freezing-Point Depression", pp721-724.

Explanation / Answer

1)

Lets calculate molality first

Molar mass of C10H8 = 10*MM(C) + 8*MM(H)

= 10*12.01 + 8*1.008

= 128.164 g/mol

mass of C10H8 = 3.95 g

we have below equation to be used:

number of mol of C10H8,

n = mass of C10H8/molar mass of C10H8

=(3.95 g)/(128.164 g/mol)

= 3.082*10^-2 mol

mass of solvent = 12.20 g

= 1.22*10^-2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(3.082*10^-2 mol)/(1.22*10^-2 Kg)

= 2.526 molal

lets now calculate deltaTf

deltaTf = Kf*m

= 5.12*2.5262

= 12.9342 oC

This is decrease in freezing point

freezing point of pure liquid = 5.5 oC

So, new freezing point = 5.5 - 12.9342

= -7.4342 oC

Answer: -7.43 oC

2)

lets now calculate deltaTb

deltaTb = Kb*m

= 2.53*2.5262

= 6.3913 oC

This is increase in boiling point

boiling point of pure liquid = 80.1 oC

So, new boiling point = 80.1 + 6.3913

= 86.49 oC

Answer: 86.49 oC

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