`3.0` molal `NaOH` solution has a density of `1.110 g//mL`. The molarity of the solution is:

Options

A. `2.9732`
B. `3.05`
C. `3.64`
D. `3.0504`

Explanation

To find the molarity of a 3.0 molal NaOH solution with a density of 1.110 g/mL, we can follow these steps:

### Step 1: Understand the Definitions
- **Molality (m)** is defined as the number of moles of solute per kilogram of solvent.
- **Molarity (M)** is defined as the number of moles of solute per liter of solution.

### Step 2: Calculate the Mass of the Solvent
Given that the solution is 3.0 molal, this means there are 3.0 moles of NaOH in 1 kg (1000 g) of solvent (water).

### Step 3: Calculate the Mass of NaOH
The molar mass of NaOH (sodium hydroxide) is approximately:
- Na: 23 g/mol
- O: 16 g/mol
- H: 1 g/mol
- Total = 23 + 16 + 1 = 40 g/mol

For 3.0 moles of NaOH:
\[ \text{Mass of NaOH} = \text{moles} \times \text{molar mass} = 3.0 \, \text{mol} \times 40 \, \text{g/mol} = 120 \, \text{g} \]

### Step 4: Calculate the Total Mass of the Solution
The total mass of the solution is the mass of the solvent plus the mass of the solute (NaOH):
\[ \text{Total mass of solution} = \text{mass of solvent} + \text{mass of NaOH} = 1000 \, \text{g} + 120 \, \text{g} = 1120 \, \text{g} \]

### Step 5: Convert the Mass of the Solution to Volume
Using the density of the solution (1.110 g/mL), we can find the volume of the solution:
\[ \text{Volume of solution} = \frac{\text{mass of solution}}{\text{density}} = \frac{1120 \, \text{g}}{1.110 \, \text{g/mL}} \approx 1009.01 \, \text{mL} \]
Convert this to liters:
\[ \text{Volume in liters} = \frac{1009.01 \, \text{mL}}{1000} \approx 1.009 \, \text{L} \]

### Step 6: Calculate the Molarity
Now, we can calculate the molarity (M) of the solution:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{3.0 \, \text{mol}}{1.009 \, \text{L}} \approx 2.97 \, \text{M} \]

### Final Answer
The molarity of the 3.0 molal NaOH solution is approximately **2.97 M**.

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