Molality $(m)$ of $3 \, \text{M}$ aqueous solution of $\text{NaCl}$ is:
(Given: Density of solution $= 1.25 \, \text{g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $\text{Na} = 23, \, \text{Cl} = 35.5$)
(Given: Density of solution $= 1.25 \, \text{g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $\text{Na} = 23, \, \text{Cl} = 35.5$)
- 2.90 m
- 2.79 m
- 1.90 m
- 3.85 m
The Correct Option is B
Approach Solution - 1
To find the molality \(m\) of a 3 M aqueous solution of NaCl, we'll use the relationship between molality and molarity. Here is the step-by-step process:
- Identify known values:
- Molarity (M) = 3 M
- Density of solution = 1.25 g/mL
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
- Calculate the mass of the solution: Since molarity (M) is given as 3 mol/L, it implies 3 moles of NaCl per 1 liter of solution.
- Mass of 3 moles of NaCl = \(3 \times 58.5 = 175.5\) g
- Total volume of the solution = 1000 mL (since we're considering 1 L for simplicity)
- Mass of solution = Volume × Density = \(1000 \, \text{mL} \times 1.25 \, \text{g/mL} = 1250 \, \text{g}\)
- Calculate the mass of water in the solution:
- Mass of water = Mass of solution - Mass of NaCl
- Mass of water = \(1250 - 175.5 = 1074.5 \, \text{g}\)
- Convert mass of water to kg for molality: \(1074.5 \, \text{g} = 1.0745 \, \text{kg}\)
- Calculate the molality: Molality \(m\) is given by the formula:
\(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)
- Molality = \(\frac{3 \, \text{moles}}{1.0745 \, \text{kg}} = 2.79 \, \text{m}\)
Thus, the molality of the solution is \(2.79\, \text{m}\).
Approach Solution -2
For a 3 M solution, 3 moles of NaCl are present in 1 liter of solution.
The formula for molality \( m \) is:
\[ \text{molality} = \frac{\text{moles of solute} \times 1000}{\text{mass of solvent in grams}} \]
Calculate the mass of the solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \times 1000 = 1250 \, \text{g} \]
Now, calculate the mass of solute (NaCl):
\[ \text{Mass of solute} = \text{moles} \times \text{molar mass} = 3 \times 58.5 = 175.5 \, \text{g} \]
Therefore, the mass of the solvent (water) is:
\[ \text{Mass of solvent} = 1250 - 175.5 = 1074.5 \, \text{g} \]
Substitute the values to find molality:
\[ \text{molality} = \frac{3 \times 1000}{1074.5} = 2.79 \, m \]
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