What will be the mass percentage of an aqueou | vedclass

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(B) Let the number of moles of $NaOH = 0.2$ and moles of $H_2O = 0.8$ (since the sum of mole fractions is $1$).
Mass of $NaOH = 	ext{moles} 	imes \

Vedclass · Accepted Answer

$35.71$

Vedclass · Answer

(B) Let the number of moles of $NaOH = 0.2$ and moles of $H_2O = 0.8$ (since the sum of mole fractions is $1$).
Mass of $NaOH = 	ext{moles} 	imes 	ext{molar mass} = 0.2 	imes 40 = 8 	ext{ g}$.
Mass of $H_2O = 	ext{moles} 	imes 	ext{molar mass} = 0.8 	imes 18 = 14.4 	ext{ g}$.
Mass percentage of $NaOH = \frac{	ext{Mass of } NaOH}{	ext{Mass of } NaOH + 	ext{Mass of } H_2O} 	imes 100$.
Mass percentage $= \frac{8}{8 + 14.4} 	imes 100 = \frac{8}{22.4} 	imes 100 \approx 35.71\%$.
Thus, option $(B)$ is correct.

What will be the mass percentage of an aqueous solution of $NaOH$ in which the mole fraction of $NaOH$ is $0.2$ (in $\% W/W$)?

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