If the density of a 5 M solution of KCl is 1 | vedclass

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(D) $5 \ M$ molar solution means there are $5 \ mol$ of $KCl$ salt in $1 \ L$ of solution.Molecular weight of $KCl = 74.55 \ g \ mol^{-1}$.Hence,the m

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$5.64$

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(D) $5 \ M$ molar solution means there are $5 \ mol$ of $KCl$ salt in $1 \ L$ of solution.Molecular weight of $KCl = 74.55 \ g \ mol^{-1}$.Hence,the mass of solute in $1 \ L$ of solution is $5 \ mol 	imes 74.55 \ g \ mol^{-1} = 372.75 \ g$.$	ext{Density} = \frac{	ext{Mass}}{	ext{Volume}}$.Mass of $1 \ L$ $(1000 \ mL)$ of solution $= 1.26 \ g \ mL^{-1} 	imes 1000 \ mL = 1260 \ g$.Mass of solvent (water) $=$ Mass of solution $-$ Mass of solute $= 1260 \ g - 372.75 \ g = 887.25 \ g = 0.88725 \ kg$.$	ext{Molality} = \frac{	ext{Number of moles of solute}}{	ext{Mass of solvent in } kg} = \frac{5 \ mol}{0.88725 \ kg} \approx 5.636 \ m$.

If the density of a $5 \ M$ solution of $KCl$ is $1.26 \ g \ mL^{-1}$,then find the molality of the solution. (molecular mass of $KCl = 74.55 \ g \ mol^{-1}$) (in $m$)

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