`0.50`g sample of impure `CaCO_3` is dissolved in 50 ml of 0.0985 (N) HCl. After the reaction is complete, the excess HCl required 6 ml of 0.105N NaOH for neutralisation. The percentage purity of `CaCO_3` in the sample is

Options

A. 42.95
B. 429.5
C. 4.295
D. 21.86

Answer

To find the percentage purity of calcium carbonate (CaCO₃) in the sample, we can follow these steps:

### Step 1: Calculate the total millimoles of HCl used
Given:
- Volume of HCl = 50 mL
- Normality of HCl = 0.0985 N

Using the formula for millimoles:
\[ \text{Millimoles of HCl} = \text{Volume (mL)} \times \text{Normality (N)} \]
\[ \text{Millimoles of HCl} = 50 \, \text{mL} \times 0.0985 \, \text{N} = 4.925 \, \text{mmol} \]

### Step 2: Calculate the millimoles of NaOH used to neutralize excess HCl
Given:
- Volume of NaOH = 6 mL
- Normality of NaOH = 0.105 N

Using the same formula:
\[ \text{Millimoles of NaOH} = \text{Volume (mL)} \times \text{Normality (N)} \]
\[ \text{Millimoles of NaOH} = 6 \, \text{mL} \times 0.105 \, \text{N} = 0.63 \, \text{mmol} \]

### Step 3: Calculate the millimoles of HCl that reacted with CaCO₃
The excess HCl is neutralized by NaOH, so the millimoles of HCl that reacted with CaCO₃ can be calculated as:
\[ \text{Millimoles of HCl reacted} = \text{Total millimoles of HCl} - \text{Millimoles of NaOH} \]
\[ \text{Millimoles of HCl reacted} = 4.925 \, \text{mmol} - 0.63 \, \text{mmol} = 4.295 \, \text{mmol} \]

### Step 4: Calculate the millimoles of CaCO₃ that reacted
The reaction between CaCO₃ and HCl is:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]
From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, the millimoles of CaCO₃ that reacted is:
\[ \text{Millimoles of CaCO₃} = \frac{\text{Millimoles of HCl reacted}}{2} = \frac{4.295 \, \text{mmol}}{2} = 2.1475 \, \text{mmol} \]

### Step 5: Convert millimoles of CaCO₃ to grams
Molar mass of CaCO₃ = 100 g/mol
\[ \text{Mass of CaCO₃} = \text{Millimoles} \times \text{Molar mass} = 2.1475 \, \text{mmol} \times \frac{100 \, \text{g}}{1000 \, \text{mmol}} = 0.21475 \, \text{g} \]

### Step 6: Calculate the percentage purity of CaCO₃
Given the initial mass of the sample is 0.50 g:
\[ \text{Percentage purity} = \left( \frac{\text{Mass of CaCO₃}}{\text{Initial mass of sample}} \right) \times 100 \]
\[ \text{Percentage purity} = \left( \frac{0.21475 \, \text{g}}{0.50 \, \text{g}} \right) \times 100 = 42.95\% \]

### Final Answer
The percentage purity of CaCO₃ in the sample is **42.95%**.

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