3.0 molal NaOH solution has a density of 1.11 | vedclass

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(D) Given,density of the solution is $1.110 \, g \, mL^{-1}$.The concentration of the $NaOH$ solution is $3.0$ molal,which means $3$ moles of $NaOH$ a

Vedclass · Accepted Answer

$2.97$

Vedclass · Answer

(D) Given,density of the solution is $1.110 \, g \, mL^{-1}$.The concentration of the $NaOH$ solution is $3.0$ molal,which means $3$ moles of $NaOH$ are present in $1000 \, g$ of solvent (water).Mass of $NaOH = 3 \, 	ext{mol} 	imes 40 \, g \, 	ext{mol}^{-1} = 120 \, g$.Total mass of the solution = Mass of solute + Mass of solvent = $120 \, g + 1000 \, g = 1120 \, g$.Volume of the solution = $\frac{	ext{Mass}}{	ext{Density}} = \frac{1120 \, g}{1.110 \, g \, mL^{-1}} \approx 1009.01 \, mL = 1.00901 \, L$.Molarity $(M)$ = $\frac{	ext{Number of moles of solute}}{	ext{Volume of solution in } L} = \frac{3 \, 	ext{mol}}{1.00901 \, L} \approx 2.97 \, M$.Therefore,the molarity of the solution is $2.97 \, M$.

$3.0$ molal $NaOH$ solution has a density of $1.110 \, g/mL$. The molarity of the solution is

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