What would be the molality of $20 %$ (mass/mass) aqueous solution of $K I$ ?
Molar mass of $K I$ is $166 g m o l^{- 1}$

 JEE MAIN 2019

1.51

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1.35

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1.08

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1.48

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Solution

The correct option is A $1.51$
$20 %$ (mass/mass) aqueous solution of $K I$ means $20 g$ of $K I$ in $100 g$ of solution.
Mass of solution =100g
Mass of solute=20g
Mass of solvent=mass of solution - mass of solute
$= (100 - 20) g = 80 g = 0.08 k g$
Molar mass of $K I$ is $166 g m o l^{- 1}$
$Number of moles of solute (n) = \frac{mass of KI}{molar mass of KI} = \frac{20}{166} = 0.121 m o l$
So, molality of solution $m = \frac{number of moles of solute}{weight of solvent (in kg)}$
$\Rightarrow m = \frac{0.121}{0.08} = 1.51 m o l / k g$
Hence, option (A) is the correct.

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What would be the molality of 20% (mass/mass) aqueous solution of KI ? Molar mass of KI is 166 g mol−1  JEE MAIN 2019

What would be the molality of $20 %$ (mass/mass) aqueous solution of $K I$ ?
Molar mass of $K I$ is $166 g m o l^{- 1}$

 JEE MAIN 2019