Calculate(a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is1.202 g mL^-1.

(a)Molar mass of KI = 39 + 127 = 166 g mol⁻¹

20%(mass/mass) aqueous solution of KI means 20g of KI is present in 100 g of solution.

That is,

20g of KI is present in (100 − 20) g of water = 80 g of water

Therefore, molality of the solution $=\frac{molesofKI}{MassofwaterinKg}$

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

$=1.506m$

$=1.51m$ (approximately)

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

∴Volume of 100 g solution $=\frac{mass}{Density}=\frac{100}{1.202}=83.19ml$

=83.19 × 10⁻³L

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

= 1.45 M

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

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

Therefore, mole fraction of KI $=\frac{molesofKI}{molesofKI+molesofwater}$

$=\frac{0.12}{0.12+4.44}=0.0263$