An aqueous KCl solution of density 1.20 g m | vedclass

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(A) Given: Molality $(m)$ = $3.30 \ mol \ kg^{-1}$,Density $(d)$ = $1.20 \ g \ mL^{-1}$,Molar mass of $KCl$ $(M_2)$ = $74.5 \ g \ mol^{-1}$.Let the ma

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$3$

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(A) Given: Molality $(m)$ = $3.30 \ mol \ kg^{-1}$,Density $(d)$ = $1.20 \ g \ mL^{-1}$,Molar mass of $KCl$ $(M_2)$ = $74.5 \ g \ mol^{-1}$.Let the mass of solvent be $1000 \ g$ $(1 \ kg)$.Moles of solute $(n_2)$ = $3.30 \ mol$.Mass of solute $(w_2)$ = $n_2 	imes M_2 = 3.30 	imes 74.5 = 245.85 \ g$.Total mass of solution $(w)$ = Mass of solvent + Mass of solute = $1000 + 245.85 = 1245.85 \ g$.Volume of solution $(V)$ = $\frac{	ext{Mass}}{	ext{Density}} = \frac{1245.85 \ g}{1.20 \ g \ mL^{-1}} = 1038.21 \ mL = 1.03821 \ L$.Molarity $(M)$ = $\frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{3.30 \ mol}{1.03821 \ L} \approx 3.178 \ mol \ L^{-1}$.The nearest integer is $3$.

An aqueous $KCl$ solution of density $1.20 \ g \ mL^{-1}$ has a molality of $3.30 \ mol \ kg^{-1}$. The molarity of the solution in $mol \ L^{-1}$ is ..... . (Nearest integer)
[Molar mass of $KCl = 74.5 \ g \ mol^{-1}$]

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An aqueous $KCl$ solution of density $1.20 \ g \ mL^{-1}$ has a molality of $3.30 \ mol \ kg^{-1}$. The molarity of the solution in $mol \ L^{-1}$ is ..... . (Nearest integer)[Molar mass of $KCl = 74.5 \ g \ mol^{-1}$]