What volume of `75%H_(2)SO_(4)` by mass is required to prepare `1.5` litres of `0.2M H_(2)SO_(4)`? (Density of the sample is `1.8 g//c c`)

Options

A. 14.2 cc
B. 28.4 cc
C. 21.7 cc
D. 7.1 cc

Explanation

To solve the problem of finding the volume of 75% H₂SO₄ required to prepare 1.5 liters of 0.2M H₂SO₄, we can follow these steps:

### Step 1: Determine the number of moles of H₂SO₄ needed
We start by calculating the number of moles of H₂SO₄ required for a 0.2M solution in 1.5 liters.

\[ \text{Moles of H₂SO₄} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of H₂SO₄} = 0.2 \, \text{mol/L} \times 1.5 \, \text{L} = 0.3 \, \text{mol} \]

### Step 2: Calculate the mass of H₂SO₄ required
Next, we convert moles of H₂SO₄ to grams using its molar mass (98 g/mol).

\[ \text{Mass of H₂SO₄} = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Mass of H₂SO₄} = 0.3 \, \text{mol} \times 98 \, \text{g/mol} = 29.4 \, \text{g} \]

### Step 3: Determine the mass of the 75% H₂SO₄ solution needed
Since the solution is 75% H₂SO₄ by mass, we can find the total mass of the solution needed to obtain 29.4 g of H₂SO₄.

\[ \text{Mass of solution} = \frac{\text{Mass of H₂SO₄}}{\text{Percentage by mass}} \]
\[ \text{Mass of solution} = \frac{29.4 \, \text{g}}{0.75} = 39.2 \, \text{g} \]

### Step 4: Calculate the volume of the 75% H₂SO₄ solution
Using the density of the 75% H₂SO₄ solution (1.8 g/cm³), we can find the volume of the solution required.

\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
\[ \text{Volume} = \frac{39.2 \, \text{g}}{1.8 \, \text{g/cm}^3} \approx 21.78 \, \text{cm}^3 \]

### Step 5: Convert the volume to milliliters
Since 1 cm³ is equivalent to 1 mL, the volume in mL is:

\[ \text{Volume} \approx 21.78 \, \text{mL} \]

### Final Answer
The volume of 75% H₂SO₄ required to prepare 1.5 liters of 0.2M H₂SO₄ is approximately **21.78 mL**.

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