Answer :
The molality of the solution, rounded to two decimal places, is 1.05 mol/kg .
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To find the molality of the solution, we use the formula:
[tex]\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \][/tex]
Given:
- - Mass of NaCl [tex](\( m_{\text{NaCl}} \))[/tex]= 151 grams
- - Mass of water[tex](\( m_{\text{water}} \))[/tex] = 2.46 kg
- Convert mass of NaCl to kilograms:**
[tex]\[ m_{\text{NaCl}} = \frac{151 \text{ grams}}{1000} = 0.151 \text{ kg} \][/tex]
- Calculate moles of NaCl:**
[tex]\[ \text{MW}_{\text{NaCl}} = 22.99 + 35.45 = 58.44 \text{ g/mol} \] \[ \text{moles of NaCl} = \frac{0.151}{58.44} \approx 2.58 \text{ moles} \][/tex]
- Calculate molality:
[tex]\[ m = \frac{2.58 \text{ moles}}{2.46 \text{ kg}} \approx 1.05 \text{ mol/kg} \][/tex]
Therefore,[tex]\( \boxed{1.05 \text{ mol/kg}} \)[/tex] is the molality of the solution.
