Find out the molality of a solution prepared | vedclass

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(A) Step $1$: Calculate the moles of solute $(NaCl)$.Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.Moles of $NaCl = \frac{2.89 \ g}{58.5 \ g/mol} \a

Vedclass · Accepted Answer

$0.311$

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(A) Step $1$: Calculate the moles of solute $(NaCl)$.Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.Moles of $NaCl = \frac{2.89 \ g}{58.5 \ g/mol} \approx 0.0494 \ mol$.Step $2$: Calculate the mass of the solvent (water).Volume of water = $0.159 \ L = 159 \ mL$.Density of water = $1.00 \ g/mL$.Mass of water = $159 \ mL 	imes 1.00 \ g/mL = 159 \ g = 0.159 \ kg$.Step $3$: Calculate molality $(m)$.$m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.0494 \ mol}{0.159 \ kg} \approx 0.311 \ m$.

Find out the molality of a solution prepared by dissolving $2.89 \ g$ of $NaCl$ in $0.159 \ L$ of water. The density of water is $1.00 \ g / mL$. (in $m$)

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