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

Step 1: Determine the mass of $KI$ and the solvent.
$20 % (mass / mass)$ aqueous solution means $20 g$ of $KI$ is dissolved in $100 g$ solution.
Therefore, mass of $KI = 20 g$
Mass of solvent $= 80 g$
Step 2: Formula for the calculation of molality
Molality $= \frac{Moles of solute}{Mass of Solvent in kg}$
Step : Calculation of molality
Molality $= \frac{Moles of solute}{Mass of Solvent in kg}$
Molality $= \frac{20 / 166}{80 / 1000}$
Molality $= 1.506 ≃ 1.51 mol / kg$
Hence, the molality of $20 % (mass / mass)$ aqueous solution of $KI$ is $1.51 mol / kg$ .

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What would be the molality of 20%(mass/mass) aqueous solution of KI? (Molar mass of KI=166gmol-1)

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