When excess of `CaCO_3` is treated with 100 ml. of HCI solution, the `CO_2` gas obtained was found to be `1.12` litre (at N.T.P.). What is the normality of HCI?

Options

A. 0.2 N
B. 1 N
C. 0.1 N
D. 2 N

Answer

To find the normality of HCl when excess CaCO₃ is treated with 100 ml of HCl solution and 1.12 liters of CO₂ gas is produced, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]

2. **Calculate Moles of CO₂ Produced:**
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Therefore, we can calculate the number of moles of CO₂ produced from the volume of CO₂ collected.
\[ \text{Moles of CO}_2 = \frac{\text{Volume of CO}_2}{\text{Molar Volume at NTP}} = \frac{1.12 \, \text{L}}{22.4 \, \text{L/mol}} = 0.05 \, \text{mol} \]

3. **Determine Moles of HCl Required:**
From the balanced equation, we see that 2 moles of HCl are required for every mole of CO₂ produced. Therefore, the moles of HCl can be calculated as:
\[ \text{Moles of HCl} = 2 \times \text{Moles of CO}_2 = 2 \times 0.05 \, \text{mol} = 0.1 \, \text{mol} \]

4. **Calculate Normality of HCl:**
Normality (N) is defined as the number of equivalents of solute per liter of solution. For HCl, which provides 1 H⁺ ion per molecule, the normality is equal to the molarity.
\[ \text{Normality} = \frac{\text{Moles of HCl}}{\text{Volume of solution in liters}} \]
The volume of HCl solution is given as 100 ml, which is equivalent to 0.1 liters (100 ml = 0.1 L).
\[ \text{Normality} = \frac{0.1 \, \text{mol}}{0.1 \, \text{L}} = 1 \, \text{N} \]

### Final Answer:
The normality of the HCl solution is **1 N**.