# Calculate (a) molality (b) molarity and (c) mole fraction of

Question:

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.

Solution:

(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 $=\frac{\text { Moles of KI }}{\text { Mass of water in } \mathrm{kg}}$

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

= 1.506 m

= 1.51 m (approximately)

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

$\therefore$ Volume of $100 \mathrm{~g}$ solution $=\frac{\text { Mass }}{\text { Density }}$

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

= 83.19 mL

= 83.19 × 10−3 L

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

= 1.45 M

(c) Moles of $\mathrm{KI}=\frac{20}{166}=0.12 \mathrm{~mol}$

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

Therefore, mole fraction of $\mathrm{KI}=\frac{\text { Moles of } \mathrm{KI}}{\text { Moles of } \mathrm{KI}+\text { Moles of water }}$

$=\frac{0.12}{0.12+4.44}$

= 0.0263