The density of an aqueous solution of 4% by m | vedclass

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(C) Given: $4\%$ $NaOH$ by mass means $4\ g$ of $NaOH$ is present in $100\ g$ of solution.Mass of solute $(NaOH)$ $= 4\ g$.Molar mass of $NaOH = 23 +

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$> 1\ m$

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(C) Given: $4\%$ $NaOH$ by mass means $4\ g$ of $NaOH$ is present in $100\ g$ of solution.Mass of solute $(NaOH)$ $= 4\ g$.Molar mass of $NaOH = 23 + 16 + 1 = 40\ g/mol$.Moles of $NaOH = \frac{4\ g}{40\ g/mol} = 0.1\ mol$.Mass of solution $= 100\ g$.Density of solution $= 1.04\ g/mL$.Volume of solution $= \frac{	ext{Mass}}{	ext{Density}} = \frac{100\ g}{1.04\ g/mL} \approx 96.15\ mL$.Mass of solvent $= 	ext{Mass of solution} - 	ext{Mass of solute} = 100\ g - 4\ g = 96\ g = 0.096\ kg$.Molality $(m)$ $= \frac{	ext{Moles of solute}}{	ext{Mass of solvent in } kg} = \frac{0.1\ mol}{0.096\ kg} \approx 1.04\ m$.Since $1.04\ m > 1\ m$,the correct option is $C$.

The density of an aqueous solution of $4\%$ by mass $NaOH$ is $1.04\ g/mL$. The molality of the solution is:

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