Molality (m) of 3 text{ M} aqueous solution o | vedclass

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(B) $3 	ext{ M}$ solution means $3 	ext{ moles}$ of $NaCl$ are present in $1 	ext{ L}$ $(1000 	ext{ mL})$ of solution.Mass of solution = $	ext{Vo

Vedclass · Accepted Answer

$2.79$

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(B) $3 	ext{ M}$ solution means $3 	ext{ moles}$ of $NaCl$ are present in $1 	ext{ L}$ $(1000 	ext{ mL})$ of solution.Mass of solution = $	ext{Volume} 	imes 	ext{Density} = 1000 	ext{ mL} 	imes 1.25 	ext{ g mL}^{-1} = 1250 	ext{ g}$.Mass of solute $(NaCl)$ = $3 	ext{ mol} 	imes 58.5 	ext{ g mol}^{-1} = 175.5 	ext{ g}$.Mass of solvent = $	ext{Mass of solution} - 	ext{Mass of solute} = 1250 	ext{ g} - 175.5 	ext{ g} = 1074.5 	ext{ g} = 1.0745 	ext{ kg}$.Molality $(m)$ = $\frac{	ext{Moles of solute}}{	ext{Mass of solvent in kg}} = \frac{3}{1.0745} = 2.79 	ext{ m}$.Alternatively, using the formula:$m = \frac{M 	imes 1000}{1000 	imes d - M 	imes M_B}$$m = \frac{3 	imes 1000}{1000 	imes 1.25 - 3 	imes 58.5} = 2.79 	ext{ m}$.

Molality $(m)$ of $3 \text{ M}$ aqueous solution of $NaCl$ is:
(Given: Density of solution = $1.25 \text{ g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $Na = 23$, $Cl = 35.5$) (in $\text{m}$)

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Molality $(m)$ of $3 \text{ M}$ aqueous solution of $NaCl$ is:(Given: Density of solution = $1.25 \text{ g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $Na = 23$, $Cl = 35.5$) (in $\text{m}$)