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  4. What would be the molality of 20 % (mass / mass) aqueous solution of KI? (molar mass of KI = 166 g mol-1)

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Q. What would be the molality of $20\%$ (mass / mass) aqueous solution of $KI$?
(molar mass of $KI = 166\, g\, mol^{-1}$)

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Solution:

$\frac{W}{W} \% = 20$
$100\, gm$ solution has $20\, gm\, KI$
$80 \,gm$ solvent has $20\, gm\, KI$
$ m = \frac{\frac{20}{166}}{\frac{80}{1000}} = \frac{20\times1000}{166\times80} = 1.506 \simeq 1.51 \,mol/kg$