Solutions and Colligative Properties - Result Question 31
2. What would be the molality of $20 %$ (mass/mass) aqueous solution of KI? (Molar mass of KI = $166 $ $g$ $ mol^{-1}$ )
(2019 Main, 9 April I)
(a) $1.48$
(b) $1.51$
(c) $1.35$
(d) $1.08$
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Answer:
Correct Answer: 2. (b)
Solution:
Key Idea: Molality is defined as number of moles of solute per kg of solvent.
$ m=\dfrac{w_2}{M w_2} \times \dfrac{1000}{w_1} $
$w_2=$ mass of solute, $M w_2=$ molecular mass of solute
$w_1=$ mass of solvent.
The molality of $20 %$ (mass $/$ mass) aqueous solution of KI can be calculated by following formula.
$ m=\dfrac{w_2 \times 1000}{M w_2 \times w_1} $
$20 %$ aqueous solution of KI means that $20$ gm of KI is present in $80$ gm solvent.
$ m=\dfrac{20}{166} \times \dfrac{1000}{80}=1.506 \approx 1.51 \mathrm{~mol} / \mathrm{kg} $
BETA
