Molality of 30 % w/w aqueous solution of NaO | vedclass

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Question

(C) Given: $30 \% \ w/w$ $NaOH$ solution.This means $30 \ g$ of $NaOH$ is present in $100 \ g$ of the solution.Mass of solute $(NaOH)$ $= 30 \ g$.Mass

Vedclass · Accepted Answer

$10.71$

Vedclass · Answer

(C) Given: $30 \% \ w/w$ $NaOH$ solution.This means $30 \ g$ of $NaOH$ is present in $100 \ g$ of the solution.Mass of solute $(NaOH)$ $= 30 \ g$.Mass of solvent (water) $= 100 \ g - 30 \ g = 70 \ g = 0.07 \ kg$.Molar mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$.Moles of $NaOH = \frac{30 \ g}{40 \ g/mol} = 0.75 \ mol$.Molality $(m)$ $= \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.75 \ mol}{0.07 \ kg} = 10.71 \ m$.

Molality of $30 \% \ w/w$ aqueous solution of $NaOH$ is . . . . . . . (in $m$)

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