The density of a $2 ext{ M}$ aqueous solution of $NaOH$ is $1.28 ext{ g/cm}^3$. The molality of the soluti

Question

The density of a $2 	ext{ M}$ aqueous solution of $NaOH$ is $1.28 	ext{ g/cm}^3$. The molality of the solution is: [molecular mass of $NaOH = 40 	ext{ g mol}^{-1}$]

Accepted Answer

1. **Understand Molarity:** A $2 	ext{ M}$ solution contains 2 moles of solute ($NaOH$) in 1 Litre ($1000 	ext{ mL}$) of solution .
2. **Calculate Mass of Solution:**
   $$ 	ext{Mass} = 	ext{Density} 	imes 	ext{Volume} $$
   $$ 	ext{Mass of solution} = 1.28 	ext{ g/mL} 	imes 1000 	ext{ mL} = 1280 	ext{ g} $$
3. **Calculate Mass of Solute ($NaOH$):**
   $$ 	ext{Mass} = 	ext{Moles} 	imes 	ext{Molar Mass} $$
   $$ 	ext{Mass of } NaOH = 2 	ext{ mol} 	imes 40 	ext{ g/mol} = 80 	ext{ g} $$
4. **Calculate Mass of Solvent (Water):**
   $$ 	ext{Mass of Solvent} = 	ext{Mass of Solution} - 	ext{Mass of Solute} $$
   $$ 	ext{Mass of Solvent} = 1280 	ext{ g} - 80 	ext{ g} = 1200 	ext{ g} = 1.2 	ext{ kg} $$
5. **Calculate Molality ($m$):**
   Molality is defined as moles of solute per kilogram of solvent .
   $$ m = \frac{	ext{Moles of Solute}}{	ext{Mass of Solvent (kg)}} $$
   $$ m = \frac{2}{1.2} \approx 1.666... 	ext{ m} $$
   Rounding to two decimal places, the molality is $1.67 	ext{ m}$.

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