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

Explanation

To solve the problem step by step, we will determine the molarity of the solution based on the given information.

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

### Step 2: Calculate the number of moles of solute
Since the molarity is defined as the number of moles of solute per liter of solution, we can express this as:
\[ \text{Number of moles of } H_2X = \text{Molarity} \times \text{Volume of solution} \]
Given that the volume of the solution is 1 liter (as per the definition of molarity):
\[ \text{Number of moles of } H_2X = 3.2 \, \text{mol/L} \times 1 \, \text{L} = 3.2 \, \text{moles} \]

### Step 3: Calculate the mass of the solvent
To find the mass of the solvent, we will use the density formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Rearranging this gives us:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
We know the density of the solvent is \( 0.4 \, \text{g/mL} \) and the volume of the solution is \( 1000 \, \text{mL} \) (1 liter):
\[ \text{Mass of solvent} = 0.4 \, \text{g/mL} \times 1000 \, \text{mL} = 400 \, \text{g} = 0.4 \, \text{kg} \]

### Step 4: Calculate the molality of the solution
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
Substituting the values we have:
\[ \text{Molality} = \frac{3.2 \, \text{moles}}{0.4 \, \text{kg}} = 8 \, \text{mol/kg} \]

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

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