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Molar mass of KI = 39 + 127 = 166 g mol-1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 - 20) g of water = 80 g of water
Therefore, molality of the solution
= moles of KI/mass of water in Kg = (20/166)/ 0.08
= 1.506 m
= 1.51 m (approximately)
(b) It is given that the density of the solution = 1.202 g mL?1
mass/ density =Volume of 100 g solution
100g/ 1.202 gml-1
= 83.19 mL
= 83.19 � 10?3 L
Therefore, molarity of the solution = (20/166) / 83.19 � 10?3 L
= 1.45 M
(c) Moles of KI
Moles of water = 80/18 = 4.44
Therefore, mole fraction of KI = moles of KI/ moles of KI + moles of water = 0.12/0.12 + 4.44
= 0.0263
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