An aqueous solution of 0 667 M nitric acid has density 1.2 g cm- Calculate
(1) Molality.
(2) Mass percentage of solute.
(3) Mole fraction of solvent
Answers
Answer:
MolaLity =
10.74
m
Molarity =
6.76
M
mole fraction =
0.162
mass of solution =
100 g
Explanation:
40% ethylene glycol means
mass of ethylene glycol =
40 g
mass of water (solvent) =
60 g = 0.060 kg
mass of solution =
100 g
1. Molality
Molar mass of ethylene glycol =
62.07 g/mL
no. of moles =
4 g
62.07 g.mol
−
1
=
0.6444 mol
Molality =
no. of moles
mass of solvent in kg
Molality =
0.6444 mol
0.060 kg
=
10.7 m
2. Molarity
Molarity
=
no. of moles
volume of solution in litres
Density
=
1.05 g/cm
3
Volume
=
Mass
Density
Volume
=
100 g
1.05 g/mL
=
95.24 mL
=
0.0952 L
Molarity
=
0.6444 mol
0.0952 L
=
6.76 M or 6.76 mol/L
3. Mole fraction
no. of moles of water
=
60 g
18.02 g.mol
-1
=
3.329 mol
Total moles
=
(3.329 + 0.6444) mol
=
3.974 mol
mole fraction of ethylene glycol
=
0.6444
3.974
=
0.162