# 2.5   Calculate(a) molality(b) molarity and(c) mole fractionof KI if the density of $20\%$ (mass/mass) aqueous KI is $\inline 1.202\; g\; mL^{-1}$.

If we assume our solution is 100 g. Then according to question,  20 g KI is present and 80 g is water.

So moles of KI   :

$=\frac{20}{166}$           $\left ( Molar\ mass = 39+127 = 166\ g\ mol^{-1} \right )$

(a) Molality :-

$Molality = \frac{Moles}{Mass\ of\ solvent\ in\ Kg} = \frac{\frac{20}{166}}{0.08}= 1.506\ m.$

(b) Molarity :-

$Density = \frac{Mass}{Volume}$

$Volume = \frac{Mass}{Density} = \frac{100}{1.202} = 83.19 mL$

$Molarity = \frac{Moles}{Volume(l)} = \frac{\frac{20}{166}}{83.19\times10^{-3}} = 1.45\ M$

(c)  Mol fraction :-   Moles of water :-

$= \frac{80}{18} = 4.44$

So, mol fraction of KI :-

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

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