What is the molality m, of methanol in a solution prepared by dissolving `160 g` of methanol, `CH_(3)OH`, in `200.0 g` of water :

Options

A. `1.0 m`
B. `5.0 m`
C. `10.0 m`
D. `25.0 m`

Explanation

To find the molality (m) of methanol in the solution, we will follow these steps:

### Step 1: Calculate the number of moles of methanol (CH₃OH)

1. **Find the molar mass of methanol (CH₃OH)**:
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol
- Oxygen (O): 16 g/mol
- Total molar mass = 12 + 4 + 16 = 32 g/mol

2. **Calculate the number of moles (n)**:
- Given mass of methanol = 160 g
- Number of moles (n) = mass (g) / molar mass (g/mol)
- n = 160 g / 32 g/mol = 5 moles

### Step 2: Calculate the mass of the solvent (water)

- Given mass of water = 200 g

### Step 3: Calculate molality (m)

- The formula for molality (m) is:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent (kg)}} \]
- Convert the mass of the solvent from grams to kilograms:
- Mass of water in kg = 200 g / 1000 = 0.2 kg

- Now substitute the values into the formula:
\[ m = \frac{5 \text{ moles}}{0.2 \text{ kg}} = 25 \text{ molal} \]

### Final Answer:
The molality of methanol in the solution is **25 molal**.

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