Given : Let mass of solution in water = 100 g, then mass of KI=20g
∴‌ Mass of solvent (water) 100−20
=80 g=0.080 kg
(a) Calculation of molalityMolar mass of KI=39+127
=166gmol−1
∴‌ Moles of KI=

20 g

166 g mol−1

=0.120
Molality of solution =

‌Number‌of‌moles‌of‌ ‌KI

‌Mass‌of‌solvent‌in‌kg

=

0.120mol

0.080 kg

=1.5 mol kg−1
(b) Calculation of molarity
Density of solution = 1.202 g mL–1
∴ Volume of solution =

100 g

1.202 gmL−1

=83.2mL=0.0832 L
Molarity of solution =

‌Number‌of‌moles‌of‌solute

Volume‌of‌solution‌in‌L

=

0.120mole

0.0832L

=1.44M
(c) Calculation of mole fraction of KI
No. of moles of KI = 0.120
No. of moles of water =

‌Mass‌of‌water

‌Molar‌mass‌of‌water

=

80 g

18 g mol−1

=4.44
Mole fraction of
=

0.120

0.120+4.44

=

0.120

4.560

=0.0263