An aqueous of diabasic acid (molecular mass = 118) containing `35.4 g` of acid per litre of the solution has density `1.0077 g mL^(-1)`.
Express the concentration in as many ways as you can?

Explanation

To express the concentration of the given aqueous solution of a dibasic acid in various ways, we will calculate molarity, molality, normality, and mole fraction step by step.

### Given Data:
- Molecular mass of the acid (M) = 118 g/mol
- Mass of acid per liter of solution = 35.4 g
- Density of solution = 1.0077 g/mL
- Volume of solution = 1 L

### Step 1: Calculate the Mass of the Solution
Using the density of the solution, we can find the mass of the solution.

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} \]
\[ \text{Mass of solution} = 1.0077 \, \text{g/mL} \times 1000 \, \text{mL} = 1007.7 \, \text{g} \]

### Step 2: Calculate the Number of Moles of the Acid
To find the number of moles of the acid, we use the formula:

\[ \text{Number of moles} = \frac{\text{Mass of solute}}{\text{Molar mass}} \]
\[ \text{Number of moles of acid} = \frac{35.4 \, \text{g}}{118 \, \text{g/mol}} = 0.3 \, \text{mol} \]

### Step 3: Calculate the Mass of the Solvent (Water)
The mass of the solvent can be calculated by subtracting the mass of the solute from the mass of the solution.

\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} \]
\[ \text{Mass of solvent} = 1007.7 \, \text{g} - 35.4 \, \text{g} = 972.3 \, \text{g} \]

### Step 4: Calculate the Number of Moles of Water
Using the molar mass of water (18 g/mol), we can find the number of moles of water.

\[ \text{Number of moles of water} = \frac{\text{Mass of solvent}}{\text{Molar mass of water}} \]
\[ \text{Number of moles of water} = \frac{972.3 \, \text{g}}{18 \, \text{g/mol}} \approx 54.01 \, \text{mol} \]

### Step 5: Calculate Molarity (M)
Molarity is defined as the number of moles of solute per liter of solution.

\[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in L}} \]
\[ \text{Molarity} = \frac{0.3 \, \text{mol}}{1 \, \text{L}} = 0.3 \, \text{M} \]

### Step 6: Calculate Molality (m)
Molality is defined as the number of moles of solute per kilogram of solvent.

\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
\[ \text{Molality} = \frac{0.3 \, \text{mol}}{0.9723 \, \text{kg}} \approx 0.308 \, \text{m} \]

### Step 7: Calculate Normality (N)
Normality is calculated as molarity multiplied by the n-factor (basicity of the acid). For a dibasic acid, n = 2.

\[ \text{Normality (N)} = \text{Molarity} \times n \]
\[ \text{Normality} = 0.3 \, \text{M} \times 2 = 0.6 \, \text{N} \]

### Step 8: Calculate Mole Fraction of Solute
The mole fraction of solute is given by the number of moles of solute divided by the total number of moles (solute + solvent).

\[ \text{Mole fraction of solute} = \frac{\text{Number of moles of solute}}{\text{Number of moles of solute} + \text{Number of moles of solvent}} \]
\[ \text{Mole fraction of solute} = \frac{0.3}{0.3 + 54.01} \approx 0.0055 \]

### Summary of Concentration Calculations:
1. **Molarity (M)**: 0.3 M
2. **Molality (m)**: 0.308 m
3. **Normality (N)**: 0.6 N
4. **Mole Fraction of Solute**: 0.0055