Ncert Exercises & Solutions

2.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.

(a) 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 $= \frac{Moles of KI}{Mass of water in kg}$

$= \frac{\frac{20}{166}}{0.08} m$

= 1.506 m

= 1.51 m (approximately)

(b) It is given that the density of the solution = 1.202 g mL⁻¹

Volume of 100 g solution $= \frac{Mass}{Density}$

$= \frac{100 g}{1.202 g {mL}^{- 1}}$

= 83.19 mL

= 83.19 × 10⁻³ L

Therefore, molarity of the solution $= \frac{\frac{20}{166} mol}{83.19 \times 10^{- 3} L}$

= 1.45 M

Moles of water $= \frac{80}{18} = 4.44 mol$

Therefore, mole fraction of KI $= \frac{Moles of KI}{Moles of KI + Moles of water} = \frac{0.12}{0.12 + 4.44} = 0.0263$