A given solution of `H_(2) SO_(4)` is labelled as 49% (w/w) , then correct statement regarding the solution is `(d = 1.3 g / ml)`

Options

A. `m = (500)/(51)`
B. `N = (1000)/(51)`
C. `% w/v = (49 xx 1.3)%`
D. M = 6.5

Answer

To solve the problem, we need to analyze the given data about the solution of sulfuric acid (H₂SO₄) that is labeled as 49% (w/w) with a density of 1.3 g/mL. We will derive various properties of the solution step by step.

### Step 1: Determine the weight of solute and solvent
Given that the solution is 49% (w/w), this means:
- Weight of solute (H₂SO₄) = 49 g
- Weight of solution = 100 g

To find the weight of the solvent:
\[ \text{Weight of solvent} = \text{Weight of solution} - \text{Weight of solute} = 100 \, \text{g} - 49 \, \text{g} = 51 \, \text{g} \]

### Step 2: Calculate the number of moles of solute
The molecular weight of H₂SO₄ is approximately 98 g/mol. To find the number of moles of H₂SO₄:
\[ \text{Number of moles} = \frac{\text{Weight of solute}}{\text{Molecular weight}} = \frac{49 \, \text{g}}{98 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 3: Calculate molality (m)
Molality is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}} = \frac{0.5 \, \text{mol}}{0.051 \, \text{kg}} \approx 9.8 \, \text{mol/kg} \]

### Step 4: Calculate the volume of the solution
Using the density of the solution (1.3 g/mL), we can find the volume of the solution:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{1.3 \, \text{g/mL}} \approx 76.92 \, \text{mL} = 0.07692 \, \text{L} \]

### Step 5: Calculate normality (N)
Normality is defined as the number of equivalents of solute per liter of solution. For H₂SO₄, it dissociates to give 2 H⁺ ions, so the equivalent factor (n-factor) is 2.
\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume of solution in L}} = \frac{\frac{49 \, \text{g}}{98 \, \text{g/mol} \times 2}}{0.07692 \, \text{L}} = \frac{0.25 \, \text{equiv}}{0.07692 \, \text{L}} \approx 3.25 \, \text{N} \]

### Step 6: Calculate molarity (M)
Molarity is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in L}} = \frac{0.5 \, \text{mol}}{0.07692 \, \text{L}} \approx 6.5 \, \text{M} \]

### Step 7: Calculate weight/volume percentage
Weight/volume percentage is calculated as:
\[ \text{Weight/Volume \%} = \frac{\text{Weight of solute (g)}}{\text{Volume of solution (mL)}} \times 100 = \frac{49 \, \text{g}}{76.92 \, \text{mL}} \times 100 \approx 63.8\% \]

### Summary of Results
- Weight of solute (H₂SO₄): 49 g
- Weight of solvent: 51 g
- Moles of solute: 0.5 mol
- Molality: 9.8 mol/kg
- Volume of solution: 0.07692 L
- Normality: 3.25 N
- Molarity: 6.5 M
- Weight/Volume %: 63.8%