The density of 2 M aqueous solution of NaOH | vedclass

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Question

(C) $2 \ M$ solution of $NaOH$ means $2 \ mol$ of $NaOH$ is present in $1 \ L$ of solution.Density of solution $= 1.28 \ g \ mL^{-1}$.Mass of solution

Vedclass · Accepted Answer

$1.67$

Vedclass · Answer

(C) $2 \ M$ solution of $NaOH$ means $2 \ mol$ of $NaOH$ is present in $1 \ L$ of solution.Density of solution $= 1.28 \ g \ mL^{-1}$.Mass of solution $=$ Volume $	imes$ Density $= 1000 \ mL 	imes 1.28 \ g \ mL^{-1} = 1280 \ g$.Mass of solute $(NaOH)$ $= 2 \ mol 	imes 40 \ g \ mol^{-1} = 80 \ g$.Mass of solvent (water) $=$ Mass of solution $-$ Mass of solute $= 1280 \ g - 80 \ g = 1200 \ g = 1.2 \ kg$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{2 \ mol}{1.2 \ kg} = 1.67 \ m$.

The density of $2 \ M$ aqueous solution of $NaOH$ is $1.28 \ g \ cm^{-3}.$ The molality of the solution is $... \ m$ [Given that molecular mass of $NaOH = 40 \ g \ mol^{-1}$]

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