The correct relationship between molality (m) | vedclass

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Question

(B) Let the volume of the solution be $1 \ L$ $(1000 \ mL)$.Then,the number of moles of solute is $M$.The mass of the solution is $d 	imes 1000 \ g$.

Vedclass · Accepted Answer

$\frac{1}{m} = \frac{d}{M} - \frac{M_{solute}}{1000}$

Vedclass · Answer

(B) Let the volume of the solution be $1 \ L$ $(1000 \ mL)$.Then,the number of moles of solute is $M$.The mass of the solution is $d 	imes 1000 \ g$.The mass of the solute is $M 	imes M_{solute} \ g$.The mass of the solvent is $(1000d - MM_{solute}) \ g$.Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.$m = \frac{M 	imes 1000}{1000d - MM_{solute}}$.Rearranging this,we get $\frac{1}{m} = \frac{1000d - MM_{solute}}{1000M} = \frac{d}{M} - \frac{M_{solute}}{1000}$.

The correct relationship between molality $(m)$ and molarity $(M)$ of a solution is ($d = $ density in $g/mL$ and $M_{solute} = $ molar mass of solute in $g/mol$)

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