3.0 molal NaOH solution has a density of 1.11 | vedclass

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(A) Given: Molality $(m)$ = $3.0 \ mol/kg$,Density $(d)$ = $1.110 \ g/mL$,Molar mass of $NaOH$ $(M_B)$ = $40 \ g/mol$.Let the mass of the solvent be $

Vedclass · Accepted Answer

$2.94$

Vedclass · Answer

(A) Given: Molality $(m)$ = $3.0 \ mol/kg$,Density $(d)$ = $1.110 \ g/mL$,Molar mass of $NaOH$ $(M_B)$ = $40 \ g/mol$.Let the mass of the solvent be $1 \ kg = 1000 \ g$.The mass of the solution = Mass of solute + Mass of solvent = $(3.0 	imes 40) + 1000 = 120 + 1000 = 1120 \ g$.The volume of the solution = $\frac{	ext{Mass of solution}}{	ext{Density}} = \frac{1120 \ g}{1.110 \ g/mL} \approx 1009.01 \ mL = 1.009 \ L$.Molarity $(M)$ = $\frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{3.0 \ mol}{1.009 \ L} \approx 2.97 \ M$.Using the formula $M = \frac{m 	imes d 	imes 1000}{1000 + (m 	imes M_B)} = \frac{3.0 	imes 1.110 	imes 1000}{1000 + (3.0 	imes 40)} = \frac{3330}{1120} \approx 2.97 \ M$.The closest option is $2.94$.

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

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