What will be the molality of 10% w/w aqueous | vedclass

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(B) Given: $10\%$ w/w aqueous solution of $NaOH$ means $10 \ g$ of $NaOH$ is present in $100 \ g$ of solution.Mass of solute $(NaOH)$ = $10 \ g$.Mass

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$2.78$

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(B) Given: $10\%$ w/w aqueous solution of $NaOH$ means $10 \ g$ of $NaOH$ is present in $100 \ g$ of solution.Mass of solute $(NaOH)$ = $10 \ g$.Mass of solvent $(H_2O)$ = $100 \ g - 10 \ g = 90 \ g = 0.09 \ kg$.Molar mass of $NaOH$ = $40 \ g \ mol^{-1}$.Number of moles of $NaOH$ = $\frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.25 \ mol}{0.09 \ kg} \approx 2.78 \ m$.

What will be the molality of $10\%$ w/w aqueous solution of $NaOH$ (in $m$)? (Molecular mass of $NaOH = 40 \ g \ mol^{-1}$)

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