6.0 g sample of `CaCO_(3)` reacts with 20 g solution of HCl having 20% by mass of HCl (density=1.10 g/mL). Calculate percentage purity of `CaCO_(3)` sample.

Options

A. `9.13 %`
B. `91.33%`
C. `54.8%`
D. `5.48%`

Explanation

To calculate the percentage purity of the \( \text{CaCO}_3 \) sample, follow these steps:

1. **Determine the amount of HCl in the solution:**

Given:
- Mass of HCl solution = 20 g
- Percentage of HCl by mass = 20%

\[ \text{Mass of HCl} = \left( \frac{20}{100} \right) \times 20 \text{ g} = 4 \text{ g} \]

2. **Calculate the volume of the HCl solution:**

Given:
- Density of HCl solution = 1.10 g/mL

\[ \text{Volume of HCl solution} = \frac{\text{Mass of HCl solution}}{\text{Density}} = \frac{20 \text{ g}}{1.10 \text{ g/mL}} = 18.18 \text{ mL} \]

3. **Calculate the normality of the HCl solution:**

Molar mass of HCl = 36.5 g/mol

\[ \text{Normality of HCl} = \frac{\text{Mass of HCl} \times 1000}{\text{Molar mass of HCl} \times \text{Volume of HCl solution}} = \frac{4 \text{ g} \times 1000}{36.5 \text{ g/mol} \times 18.18 \text{ mL}} = 6.06 \text{ N} \]

4. **Calculate the milliequivalents of HCl:**

\[ \text{Milliequivalents of HCl} = \text{Normality of HCl} \times \text{Volume of HCl solution} = 6.06 \times 18.18 = 110.1 \text{ meq} \]

5. **Determine the milliequivalents of \( \text{CaCO}_3 \):**

Since the reaction between \( \text{CaCO}_3 \) and HCl is 1:2 (one mole of \( \text{CaCO}_3 \) reacts with two moles of HCl),

\[ \text{Milliequivalents of } \text{CaCO}_3 = \frac{\text{Milliequivalents of HCl}}{2} = \frac{110.1}{2} = 55.05 \text{ meq} \]

6. **Calculate the mass of pure \( \text{CaCO}_3 \):**

Equivalent weight of \( \text{CaCO}_3 \) = \frac{\text{Molar mass of } \text{CaCO}_3}{2} = \frac{100 \text{ g/mol}}{2} = 50 \text{ g/eq}

\[ \text{Mass of pure } \text{CaCO}_3 = \frac{\text{Milliequivalents of } \text{CaCO}_3 \times \text{Equivalent weight of } \text{CaCO}_3}{1000} = \frac{55.05 \times 50}{1000} = 2.75 \text{ g} \]

7. **Calculate the percentage purity of the \( \text{CaCO}_3 \) sample:**

Given:
- Mass of \( \text{CaCO}_3 \) sample = 6 g

\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure } \text{CaCO}_3}{\text{Mass of } \text{CaCO}_3 \text{ sample}} \right) \times 100 = \left( \frac{2.75 \text{ g}}{6 \text{ g}} \right) \times 100 = 45.83\% \]