CBSE Class 11-science Answered

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

= 0.0263