The density of a 3.6 `M H_(2)SO_(4)` solution that is 29% `H_(2)SO_(4)` by mass will be

Options

A. `0.212 gmML^(-2)`
B. `0.122 gm Ml^(-1)`
C. `2.12 gm Ml^(-1)`
D. `1.22 gm Ml^(-1)`

Answer

To find the density of a 3.6 M H₂SO₄ solution that is 29% H₂SO₄ by mass, we can follow these steps:

### Step 1: Understand the Molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. A 3.6 M solution means that in 1 liter (1000 mL) of the solution, there are 3.6 moles of H₂SO₄.

### Step 2: Calculate the Mass of H₂SO₄
To find the mass of H₂SO₄ in the solution, we need to know the molar mass of H₂SO₄. The molar mass of H₂SO₄ is calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- S: 32 g/mol × 1 = 32 g/mol
- O: 16 g/mol × 4 = 64 g/mol

Adding these together:
\[ \text{Molar mass of H₂SO₄} = 2 + 32 + 64 = 98 \text{ g/mol} \]

Now, calculate the mass of H₂SO₄ in 3.6 moles:
\[ \text{Mass of H₂SO₄} = 3.6 \text{ moles} \times 98 \text{ g/mol} = 352.8 \text{ g} \]

### Step 3: Set Up the Mass of the Solution
Let the density of the solution be \( d \) g/mL. The mass of 1 liter (1000 mL) of the solution is:
\[ \text{Mass of solution} = 1000 \text{ mL} \times d \text{ g/mL} = 1000d \text{ g} \]

### Step 4: Calculate the Percentage by Mass
The percentage by mass of H₂SO₄ in the solution is given by the formula:
\[ \text{Percentage by mass} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100 \]

Substituting the values we have:
\[ 29 = \left( \frac{352.8}{1000d} \right) \times 100 \]

### Step 5: Solve for Density \( d \)
Rearranging the equation to solve for \( d \):
\[ 29 = \frac{352.8 \times 100}{1000d} \]

This simplifies to:
\[ 29d = \frac{35280}{1000} \]
\[ 29d = 35.28 \]

Now, divide both sides by 29:
\[ d = \frac{35.28}{29} \]
\[ d \approx 1.22 \text{ g/mL} \]

### Final Answer
The density of the 3.6 M H₂SO₄ solution is approximately **1.22 g/mL**.

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