If the density of methanol is `0.8` kg `L^(-1)`, what is its volume needed for making 4 L of its `0.25` M solution?

Options

A. 4 mL
B. 8 mL
C. 40 mL
D. 80 mL

Explanation

To determine the volume of methanol needed to prepare 4 liters of a 0.25 M solution, we can follow these steps:

### Step 1: Understand the Molarity Formula
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for molarity is:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

### Step 2: Calculate the Moles of Methanol Required
We need to find the number of moles of methanol required to make a 0.25 M solution in a total volume of 4 L. Using the molarity formula:
\[ \text{moles of methanol} = M \times \text{volume of solution} \]
Substituting the values:
\[ \text{moles of methanol} = 0.25 \, \text{mol/L} \times 4 \, \text{L} = 1 \, \text{mol} \]

### Step 3: Find the Molar Mass of Methanol
The molar mass of methanol (CH₃OH) can be calculated as follows:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol × 4 = 4.032 g/mol
- Oxygen (O): 16.00 g/mol

Adding these together:
\[ \text{Molar mass of methanol} = 12.01 + 4.032 + 16.00 = 32.042 \, \text{g/mol} \]

### Step 4: Calculate the Mass of Methanol Required
Now we can calculate the mass of methanol needed using the number of moles and the molar mass:
\[ \text{mass of methanol} = \text{moles} \times \text{molar mass} \]
Substituting the values:
\[ \text{mass of methanol} = 1 \, \text{mol} \times 32.042 \, \text{g/mol} = 32.042 \, \text{g} \]

### Step 5: Convert Mass to Volume Using Density
We know the density of methanol is 0.8 kg/L, which is equivalent to 0.8 g/mL. We can use the density to find the volume:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \]
Substituting the values:
\[ \text{Volume} = \frac{32.042 \, \text{g}}{0.8 \, \text{g/mL}} = 40.0525 \, \text{mL} \]

### Step 6: Final Answer
The volume of methanol needed to prepare 4 liters of a 0.25 M solution is approximately:
\[ \text{Volume} \approx 40.05 \, \text{mL} \]