When 3 g of acetic acid is added to 500 g of water, the freezing point of solution is observed to be 0.24c. The degree of ionisation of acetic acid is (kf (water) = 1.8 k kg mol1)?
Answers
Answer:
You use the formula for freezing point depression.
Explanation:
EXAMPLE
What is the freezing point depression caused by adding 31.65 g of sodium chloride to 220.0 g of water.
K
f
for water is
1.86 °C⋅kg⋅mol
-1
.
Solution
The formula for freezing point depression expression is
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
Δ
T
f
=
i
K
f
m
a
a
∣
∣
−−−−−−−−−−−−−−−
where
Δ
T
f
is the freezing point depression,
i
is the van’t Hoff factor,
K
f
is the molal freezing point depression constant for the solvent, and
m
is the molality of the solution.
Step 1: Calculate the molality of the NaCl
moles of NaCl
=
31.65
g NaCl
×
1
mol NaCl
58.44
g NaCl
=
0.5416 mol NaCl
mass of water
=
220.0
g H
2
O
×
1
kg H
2
O
1000
g H
2
O
=
0.220 kg H
2
O
m
=
moles of NaCl
kilograms of water
=
0.5416 mol
0.220 kg
=
2.46 mol/kg
Step 2: Determine the van't Hoff factor
The van't Hoff factor,
i
, is the number of moles of particles obtained when 1 mol of a solute dissolves.
Nonelectrolytes such as sugar do not dissociate in water. One mole of solid sugar gives one mole of dissolved sugar molecules.
For nonelectrolytes,
i
=
1
.
Electrolytes such as
NaCl
completely dissociate into ions.
NaCl
→
Na
+
+
Cl
-
One mole of solid
NaCl
gives two moles of dissolved particles: 1 mol of
Na
+
ions and 1 mol of
Cl
-
ions. Thus, for
NaCl
,
i
=
2
.
Step 3: Calculate
Δ
T
f
Δ
T
f
=
i
K
f
m
=
2
×
1.86 °C⋅
kg⋅mol
-1
×
2.46
mol⋅kg
-1
=
9.16 °C
The freezing point depression is 9.16 °C.