Answer :
Answer:
Let's tackle the problem step-by-step to find the molarity of the concentrated acid solution and the volume needed to prepare the 0.2000 M dilute solution.
### Step 1: Calculate the Molarity of the Concentrated Acid Solution
Given:
- Density of the concentrated sulphuric acid (\(\rho\)) = 1.804 g/cm³
- Purity = 98%
- Molar mass of \( \text{H}_2\text{SO}_4 \) = 98 g/mol
First, convert the density to g/L:
\[ \text{Density} = 1.804 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3/\text{L} = 1804 \, \text{g/L} \]
Calculate the mass of \( \text{H}_2\text{SO}_4 \) in 1 L:
\[ \text{Mass of pure } \text{H}_2\text{SO}_4 = 1804 \, \text{g/L} \times 0.98 = 1767.92 \, \text{g/L} \]
Now, find the molarity (M) of the concentrated acid:
\[ \text{Molarity} = \frac{\text{Mass of } \text{H}_2\text{SO}_4}{\text{Molar mass of } \text{H}_2\text{SO}_4} = \frac{1767.92 \, \text{g/L}}{98 \, \text{g/mol}} = 18.04 \, \text{M} \]
So, the molarity of the concentrated \( \text{H}_2\text{SO}_4 \) is 18.04 M.
### Step 2: Calculate the Volume of Concentrated Acid Needed to Prepare 1 L of 0.2000 M Dilute Solution
We need to dilute the concentrated acid to get a solution of 0.2000 M. We use the dilution formula:
\[ M_1V_1 = M_2V_2 \]
where:
- \( M_1 \) = molarity of concentrated acid = 18.04 M
- \( V_1 \) = volume of concentrated acid needed
- \( M_2 \) = molarity of dilute solution = 0.2000 M
- \( V_2 \) = volume of dilute solution = 1 L = 1000 mL
Rearranging the formula to solve for \( V_1 \):
\[ V_1 = \frac{M_2V_2}{M_1} = \frac{0.2000 \, \text{M} \times 1000 \, \text{mL}}{18.04 \, \text{M}} \approx 11.09 \, \text{mL} \]
Therefore, approximately 11.09 mL of the concentrated \( \text{H}_2\text{SO}_4 \) is needed to prepare 1 L of 0.2000 M \( \text{H}_2\text{SO}_4 \).
### Step 3: Calculate the Molarity of the Acid Solution Prepared by the Student
The student measured 10.00 mL of the concentrated acid and diluted it to 1000 mL. Using the dilution formula again:
Given:
- \( M_1 \) = 18.04 M (molarity of concentrated acid)
- \( V_1 \) = 10.00 mL
- \( V_2 \) = 1000.00 mL
We need to find \( M_2 \):
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{18.04 \, \text{M} \times 10.00 \, \text{mL}}{1000.00 \, \text{mL}} = 0.1804 \, \text{M} \]
So, the molarity of the acid solution prepared by the student is 0.1804 M.
