The density of a 3 , M , NaCl solution is 1.2 | vedclass

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Question

(A) Given: Molarity $(M)$ = $3.0 \, mol \, L^{-1}$,Density $(d)$ = $1.25 \, g \, mL^{-1}$.In $1 \, L$ of solution,the number of moles of $NaCl$ is $3

Vedclass · Accepted Answer

$2.79$

Vedclass · Answer

(A) Given: Molarity $(M)$ = $3.0 \, mol \, L^{-1}$,Density $(d)$ = $1.25 \, g \, mL^{-1}$.In $1 \, L$ of solution,the number of moles of $NaCl$ is $3 \, mol$.Mass of $NaCl$ = $3 \, mol 	imes 58.5 \, g \, mol^{-1} = 175.5 \, g$.Mass of $1 \, L$ solution = $Volume 	imes Density = 1000 \, mL 	imes 1.25 \, g \, mL^{-1} = 1250 \, g$.Mass of solvent (water) = $Mass \, of \, solution - Mass \, of \, solute = 1250 \, g - 175.5 \, g = 1074.5 \, g = 1.0745 \, kg$.Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{3 \, mol}{1.0745 \, kg} \approx 2.79 \, m$.

The density of a $3 \, M \, NaCl$ solution is $1.25 \, g \, mL^{-1}$. Calculate the molality of the solution. (in $, m$)

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