Concentrated aqueous sulphuric acid is 98% `H_(2)SO_(4) ` by weight and has a density of `1.80 gmL^(-1)`. Molarity of solution

Options

A. 1 M
B. 1.8 M
C. 18 M
D. 1.5 M

Answer

To find the molarity of concentrated aqueous sulfuric acid (H₂SO₄) that is 98% by weight and has a density of 1.80 g/mL, we can follow these steps:

### Step 1: Determine the mass of H₂SO₄ in 100 g of solution
Since the solution is 98% H₂SO₄ by weight, in 100 g of solution, the mass of H₂SO₄ is:
\[ \text{Mass of H₂SO₄} = 98\% \text{ of } 100 \, \text{g} = 98 \, \text{g} \]

### Step 2: Calculate the volume of the solution
Using the density of the solution, we can find the volume. The density (d) is given as 1.80 g/mL. We can use the formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Rearranging this gives:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{100 \, \text{g}}{1.80 \, \text{g/mL}} \approx 55.56 \, \text{mL} \]

### Step 3: Convert the volume from mL to L
To convert the volume from mL to L:
\[ \text{Volume in L} = \frac{55.56 \, \text{mL}}{1000} = 0.05556 \, \text{L} \]

### Step 4: Calculate the number of moles of H₂SO₄
To find the number of moles, we use the molar mass of H₂SO₄. The molar mass is calculated as follows:
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Sulfur (S): 32 g/mol × 1 = 32 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Thus, the molar mass of H₂SO₄ is:
\[ \text{Molar mass of H₂SO₄} = 2 + 32 + 64 = 98 \, \text{g/mol} \]

Now, we can calculate the number of moles:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{98 \, \text{g}}{98 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 5: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{1 \, \text{mol}}{0.05556 \, \text{L}} \approx 18 \, \text{M} \]

### Final Answer
The molarity of the concentrated aqueous sulfuric acid solution is approximately **18 M**.

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