The density of `3M` solution of `NaCl` is `1.25 g mL^(-1)`. The molality of the solution is

Options

A. `3.86 mol kg^(-1)`
B. `1.97 mol kg^(-1)`
C. `2.79 mol kg^(-1)`
D. `0.786 mol kg^(-1)`

Explanation

To find the molality of a 3M NaCl solution with a density of 1.25 g/mL, follow these steps:

### Step 1: Calculate the mass of the solution
Given:
- Molarity (M) = 3 M
- Density (ρ) = 1.25 g/mL
- Volume of solution (V) = 1 L = 1000 mL

The mass of the solution can be calculated using the formula:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of solution} = 1.25 \, \text{g/mL} \times 1000 \, \text{mL} \]
\[ \text{Mass of solution} = 1250 \, \text{g} \]

### Step 2: Calculate the mass of NaCl in the solution
Given:
- Molarity (M) = 3 M
- Volume of solution (V) = 1 L = 1000 mL

Number of moles of NaCl in 1 L of solution:
\[ \text{Number of moles of NaCl} = 3 \, \text{moles} \]

Molecular weight of NaCl (Na = 23, Cl = 35.5):
\[ \text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol} \]

Mass of NaCl:
\[ \text{Mass of NaCl} = \text{Number of moles} \times \text{Molecular weight} \]
\[ \text{Mass of NaCl} = 3 \, \text{moles} \times 58.5 \, \text{g/mol} \]
\[ \text{Mass of NaCl} = 175.5 \, \text{g} \]

### Step 3: Calculate the mass of the solvent (water)
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute (NaCl)} \]
\[ \text{Mass of solvent} = 1250 \, \text{g} - 175.5 \, \text{g} \]
\[ \text{Mass of solvent} = 1074.5 \, \text{g} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
\[ \text{Mass of solvent in kg} = 1074.5 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 1.0745 \, \text{kg} \]

\[ \text{Molality (m)} = \frac{3 \, \text{moles}}{1.0745 \, \text{kg}} \]
\[ \text{Molality (m)} \approx 2.79 \, \text{mol/kg} \]

Thus, the molality of the solution is approximately \(2.79 \, \text{mol/kg}\).