Concise Physical Chemistry

c12 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come

198 SOLUTION CHEMISTRY

Exercise 12.1

(a) Calculate the mole fractions of solute and solvent in a 1.000 molar solution

of sucrose in water.

(b) Calculate the mole fractions of solute and solvent in a 1.000 molal solution

of sucrose in water.

Solution 12.1

(a) The number of grams of sucrose C 12 H 22 O 11 (M=342.3) in 1.000 liter

of the solution is 342. 3 / 1 .000 dm^3. The densityρ=m/V of pure solid

sucrose is 1.580 g cm−^3 so sucrose takes upV=m/ρ= 342. 3 / 1. 580 =

216 .6cm^3 , leaving 783.4 cm^3 of water in the 1.000 dm^3 of solution. We

have 783.4/18.02=43.474 moles of water. The mole fractions are

X 2 =

n 2

n 1 +n 2

=

1. 000

43. 474 + 1. 000

= 0. 02248

and

X 1 =

n 1

n 1 +n 2

=

43. 474

43. 474 + 0. 100

= 0. 9775

for a total of 1.000.

Comment: These mole fractions depend on the assumption that the volume that su-

crose molecules take up in water is the same as the molecular volume in the crystalline

state. This is extremely unlikely, meaning that our mole fractions calculated in this

way are probably wrong.

(b) The mole fractions of the 1.000 molal solution are

X 2 =

n 2

n 1 +n 2

=

1. 000

55. 494 + 1. 000

= 0. 0177

and

X 1 =

n 1

n 1 +n 2

=

55. 494

55. 494 + 1. 000

= 0. 9823

With a sum of 1.000 as before. Under these circumstances, the 1.000 molar

and 1.000 molal solutions are very different.

Comment: There are no assumptions concerning the volume of sucrose molecules in

solution therefore the molality is correct as written.