If 100 g of a 0.2 m aqueous solution is dil | vedclass

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(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Initial solution: $100 \ g$ of $0.2 \ m$ aqueous solution. Let

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$0.2/3 \ m$

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(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Initial solution: $100 \ g$ of $0.2 \ m$ aqueous solution. Let the mass of solute be $w_2 \ g$ and molar mass be $M_2$. $0.2 = \frac{w_2 / M_2}{mass \ of \ solvent \ in \ kg}$. Since the solution is dilute, we can approximate the mass of the solvent as $100 \ g = 0.1 \ kg$. Thus, moles of solute $n_2 = 0.2 	imes 0.1 = 0.02 \ mol$. After adding $200 \ g$ of water, the new mass of the solvent is $100 \ g + 200 \ g = 300 \ g = 0.3 \ kg$. The moles of solute remain constant at $0.02 \ mol$. New molality $m' = \frac{0.02 \ mol}{0.3 \ kg} = \frac{0.02}{0.3} = \frac{2}{30} = \frac{1}{15} \ m$. Wait, re-evaluating: $0.2 \ m$ means $0.2 \ mol$ of solute in $1000 \ g$ of solvent. In $100 \ g$ of solution, if we assume density $\approx 1 \ g/mL$, solvent mass $\approx 100 \ g$. Moles of solute $= 0.2 	imes 0.1 = 0.02 \ mol$. New solvent mass $= 100 \ g + 200 \ g = 300 \ g = 0.3 \ kg$. New molality $= \frac{0.02}{0.3} = \frac{2}{30} = 0.066 \ m$. Looking at the options, $0.2/3 \ m$ is $0.066 \ m$.

If $100 \ g$ of a $0.2 \ m$ aqueous solution is diluted by adding $200 \ g$ of water, what will be the molality of the resulting solution?

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