The density of 2 M MgBr_2 solution is 1.2 | vedclass

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(B) Given: Molarity $(M)$ = $2 \ M$,Density $(d)$ = $1.2 \ g/mL$. The molar mass of $MgBr_2$ is $24 + 2 	imes 80 = 184 \ g/mol$. Let the volume of th

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$2.4$

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(B) Given: Molarity $(M)$ = $2 \ M$,Density $(d)$ = $1.2 \ g/mL$. The molar mass of $MgBr_2$ is $24 + 2 	imes 80 = 184 \ g/mol$. Let the volume of the solution be $1 \ L = 1000 \ mL$. Mass of the solution = $Density 	imes Volume = 1.2 \ g/mL 	imes 1000 \ mL = 1200 \ g$. Moles of solute $(MgBr_2)$ = $Molarity 	imes Volume(L) = 2 \ mol/L 	imes 1 \ L = 2 \ mol$. Mass of solute = $Moles 	imes Molar \ mass = 2 \ mol 	imes 184 \ g/mol = 368 \ g$. Mass of solvent = $Mass \ of \ solution - Mass \ of \ solute = 1200 \ g - 368 \ g = 832 \ g = 0.832 \ kg$. Molality $(m)$ = $\frac{Moles \ of \ solute}{Mass \ of \ solvent \ in \ kg} = \frac{2 \ mol}{0.832 \ kg} \approx 2.403 \ m$. Thus,the molality is $2.4 \ m$.

The density of $2 \ M \ MgBr_2$ solution is $1.2 \ g/mL$. Then the molality of the solution is ........ $m$.

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