The Total solution for NCERT class 6-12
Calculate (a) molality (b) molarityand (c) mole fraction of KI if the density of 20% (mass/mass)aqueous KI is 1.202 g mL-1.
(a) Molarmass 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
= 1.506 m
= 1.51 m (approximately)
(b) Itis given that the density of the solution = 1.202 g mL−1
= 83.19 mL
= 83.19 × 10−3 L
= 1.45 M
= 0.0263