Question

Given that the vapor pressure of water is 17.54 Torr at 20 degrees celcius, calculate the vapor pressure lowering of aqueous solutions that are 1.70 m in (a) sucrose C12H22O11 and (b) Aluminum Chloride. Assume 100% dissosiation for electrolytes.

Answer 1

hello !!

given that the solution is 1.7 molal
this means 1.7 moles of solute is dissolved in 1000 g of water

for sucrose
moles of solute ;Nb = 1.7 moles

moles of H2O ;Na= 1000g/18 g/mol= 55.55 moles

Now calculate molefraction of solvent water
Xa= Na/Na+Nb = 5.55mols/[5.55moles + 1.7 moles] = 0.97 ***

Use this data while solving

A)

Use Raoult's law pressure of solution

p(solution) = mole fraction of solvent*p(solvent)

=0.97 * 17.54 torr = 17.02 torr

lowering of vapor pressure= Vapor pressure of solvent- vapor pressure of solution
DeltaP = 17.54 torr - 17.02 torr = 0.52 Torr
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B) AlCl3 ionize completely to give 4 ions in water
AlCl3 ---> Al3+ + 3Cl-

So vant hoff factor i = 4
This number has to be considered while calculating mole fraction of solvent

since Nb= 4*1.7 moles
[i was equal to 1 in the case of sucrose since it was a non electrolyte
but here AlCl3 is an electrolyte hence we have to consider i value which
depends on number of ions]

molefraction of solvent water
Xa=[ Na/(Na+(Nb*i)] = 55.55mols/[55.55moles + 1.7*4 moles] =0.89 mole

p(solution) = mole fraction of solvent*p(solvent)

p(soln) = 0.89*17.54 =15.62 torr

lowering of vapor pressure deltaP = Vapor pressure of solvent- vapor pressure of solution

delta P = 17.54 torr-15.62 torr = 1.92 torr

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THank you .hope it is helpful