A 6.50 molal solution of KOH (aq.) has a dens | vedclass

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(C) Molality $(m)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.Given $6.50$ molal $KOH$ solution means $6.50 \ moles$ of $KO

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$9$

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(C) Molality $(m)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.Given $6.50$ molal $KOH$ solution means $6.50 \ moles$ of $KOH$ are present in $1000 \ g$ of water.Molar mass of $KOH = 39.0 + 16.0 + 1.0 = 56.0 \ g \ mol^{-1}$.Mass of solute $(KOH)$ $= 6.50 \ mol 	imes 56.0 \ g \ mol^{-1} = 364.0 \ g$.Total mass of solution $= 	ext{Mass of solvent} + 	ext{Mass of solute} = 1000 \ g + 364.0 \ g = 1364.0 \ g$.Density of solution $= 1.89 \ g \ cm^{-3}$.Volume of solution $= \frac{	ext{Mass}}{	ext{Density}} = \frac{1364.0 \ g}{1.89 \ g \ cm^{-3}} \approx 721.69 \ cm^3 = 0.72169 \ L$.Molarity $(M)$ $= \frac{	ext{moles of solute}}{	ext{Volume of solution in } L} = \frac{6.50 \ mol}{0.72169 \ L} \approx 9.006 \ mol \ L^{-1}$.Rounding to the nearest integer,the molarity is $9 \ mol \ dm^{-3}$.

$A$ $6.50$ molal solution of $KOH$ $(aq.)$ has a density of $1.89 \ g \ cm^{-3}$. The molarity of the solution is .......... $mol \ dm^{-3}$.
(Round off to the Nearest Integer).
[Atomic masses: $K = 39.0 \ u, O = 16.0 \ u, H = 1.0 \ u$]

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$A$ $6.50$ molal solution of $KOH$ $(aq.)$ has a density of $1.89 \ g \ cm^{-3}$. The molarity of the solution is .......... $mol \ dm^{-3}$.(Round off to the Nearest Integer).[Atomic masses: $K = 39.0 \ u, O = 16.0 \ u, H = 1.0 \ u$]