The molality of 1 M solution of NaCl (specif | vedclass

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(B) Given: Molarity $(M)$ = $1 \ M$,Density $(d)$ = $1.0585 \ g/mL$,Molar mass of $NaCl$ = $58.5 \ g/mol$.Consider $1 \ L$ $(1000 \ mL)$ of solution.M

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(B) Given: Molarity $(M)$ = $1 \ M$,Density $(d)$ = $1.0585 \ g/mL$,Molar mass of $NaCl$ = $58.5 \ g/mol$.Consider $1 \ L$ $(1000 \ mL)$ of solution.Moles of $NaCl$ = $1 \ mol$.Mass of $NaCl$ = $1 \ mol 	imes 58.5 \ g/mol = 58.5 \ g$.Mass of solution = $	ext{Volume} 	imes 	ext{Density} = 1000 \ mL 	imes 1.0585 \ g/mL = 1058.5 \ g$.Mass of solvent $(H_2O)$ = $	ext{Mass of solution} - 	ext{Mass of solute} = 1058.5 \ g - 58.5 \ g = 1000 \ g = 1 \ kg$.Molality $(m)$ = $\frac{	ext{Moles of solute}}{	ext{Mass of solvent in } kg} = \frac{1 \ mol}{1 \ kg} = 1 \ m$.

The molality of $1 \ M$ solution of $NaCl$ (specific gravity $1.0585 \ g/mL$) is

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