What would be the molality of 20% (mass/mass) | vedclass

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(C) $20\% \, w/w$ aqueous solution of $KI$ means $20 \, g$ of $KI$ is present in $100 \, g$ of solution.Mass of solute $(KI) = 20 \, g$.Mass of solven

Vedclass · Accepted Answer

$1.51$

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(C) $20\% \, w/w$ aqueous solution of $KI$ means $20 \, g$ of $KI$ is present in $100 \, g$ of solution.Mass of solute $(KI) = 20 \, g$.Mass of solvent (water) $= 100 \, g - 20 \, g = 80 \, g = 0.08 \, kg$.Molar mass of $KI = 166 \, g \, mol^{-1}$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg}$.Moles of $KI = \frac{20 \, g}{166 \, g \, mol^{-1}} \approx 0.1205 \, mol$.Molality $(m) = \frac{0.1205 \, mol}{0.08 \, kg} = 1.506 \, mol \, kg^{-1} \approx 1.51 \, m$.

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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