Density of a solution containing `x%` by mass of `H_(2)SO_(4)` is y. The normality is

Options

A. `(xyxx10)/(98)`
B. `(xyxx10)/(98y)xx2`
C. `(xyxx10)/(98)xx2`
D. `(x xx10)/(98y)`

Explanation

To find the normality of a solution containing `x%` by mass of `H₂SO₄` with a density of `y`, we can follow these steps:

### Step 1: Understand the Definitions
- **Normality (N)** is defined as the number of equivalents of solute per liter of solution.
- The formula for normality is given by:
\[ N = \frac{\text{mass of solute (g)}}{\text{equivalent weight of solute (g/equiv)} \times \text{volume of solution (L)}} \]

### Step 2: Calculate the Mass of Solute
- In a solution with `x%` by mass of `H₂SO₄`, this means that in 100 g of the solution, there are `x g` of `H₂SO₄`.

### Step 3: Calculate the Volume of the Solution
- The density of the solution is given as `y g/mL`.
- The volume of the solution can be calculated using the formula:
\[ \text{Volume (mL)} = \frac{\text{mass of solution (g)}}{\text{density (g/mL)}} \]
- For 100 g of solution, the volume is:
\[ V = \frac{100 \text{ g}}{y \text{ g/mL}} = \frac{100}{y} \text{ mL} \]
- Convert this volume to liters:
\[ V = \frac{100}{y \times 1000} \text{ L} = \frac{100}{1000y} = \frac{1}{10y} \text{ L} \]

### Step 4: Calculate the Equivalent Weight of H₂SO₄
- The molecular weight of `H₂SO₄` is 98 g/mol.
- The n-factor for `H₂SO₄` (which can donate 2 H⁺ ions) is 2.
- Therefore, the equivalent weight (E) is:
\[ E = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{98 \text{ g/mol}}{2} = 49 \text{ g/equiv} \]

### Step 5: Substitute Values into the Normality Formula
- Now we can substitute the values into the normality formula:
\[ N = \frac{x \text{ g}}{49 \text{ g/equiv} \times \frac{1}{10y} \text{ L}} \]
- This simplifies to:
\[ N = \frac{x \times 10y}{49} \]

### Final Result
Thus, the normality of the `H₂SO₄` solution is:
\[ N = \frac{10xy}{49} \]