A compound `H_(2)X` with molar mass of `80 g` is dissolved in a solvent having density of `0.4 g mL^(-1)`. Assuming no change in volume upon dissolution, the molality of a `3.2` molar solution is

Options

A. `9`
B. `6`
C. `5`
D. `8`

Explanation

To find the molality of a 3.2 molar solution of the compound \( H_2X \) with a molar mass of 80 g/mol, we will follow these steps:

### Step 1: Understand the given data
- Molar mass of \( H_2X = 80 \, \text{g/mol} \)
- Molarity of the solution = 3.2 M
- Density of the solvent = 0.4 g/mL

### Step 2: Calculate the mass of the solvent
Since we are assuming no change in volume upon dissolution, we can use the molarity to find the volume of the solution.

1. Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Given that the molarity is 3.2 M, we can assume that we have 3.2 moles of \( H_2X \) in 1 liter (1000 mL) of solution.

### Step 3: Calculate the mass of the solvent
2. Since the density of the solvent is 0.4 g/mL, the mass of the solvent can be calculated as:
\[ \text{Mass of solvent} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of solvent} = 0.4 \, \text{g/mL} \times 1000 \, \text{mL} = 400 \, \text{g} \]

### Step 4: Calculate the molality
3. Molality (m) is defined as:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Converting the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent in kg} = \frac{400 \, \text{g}}{1000} = 0.4 \, \text{kg} \]
Now, substituting the values into the molality formula:
\[ \text{Molality} = \frac{3.2 \, \text{moles}}{0.4 \, \text{kg}} = 8 \, \text{mol/kg} \]

### Conclusion
The molality of the solution is \( 8 \, \text{mol/kg} \).