One gram limestone is heated and quickline so formed is dissolved in one litre of water. The normality of solution is `0.01 xx x`.What is value of x.

Answer

To solve the problem step-by-step, we will follow the chemical reactions and calculations involved in the process of heating limestone and dissolving quicklime in water.

### Step 1: Understand the Reaction
When limestone (calcium carbonate, CaCO₃) is heated, it decomposes into quicklime (calcium oxide, CaO) and carbon dioxide (CO₂):
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 2: Calculate the Molar Mass of Calcium Carbonate (CaCO₃)
The molar mass of calcium carbonate is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are 3 oxygen atoms)

Thus, the molar mass of CaCO₃:
\[ \text{Molar mass of CaCO}_3 = 40 + 12 + (16 \times 3) = 100 \text{ g/mol} \]

### Step 3: Calculate the Molar Mass of Calcium Oxide (CaO)
The molar mass of calcium oxide is:
- Calcium (Ca) = 40 g/mol
- Oxygen (O) = 16 g/mol

Thus, the molar mass of CaO:
\[ \text{Molar mass of CaO} = 40 + 16 = 56 \text{ g/mol} \]

### Step 4: Determine the Amount of Quicklime Produced
From the reaction, 100 g of CaCO₃ produces 56 g of CaO. Therefore, if we start with 1 g of CaCO₃:
\[ \text{Quicklime produced} = \left(\frac{56 \text{ g CaO}}{100 \text{ g CaCO}_3}\right) \times 1 \text{ g CaCO}_3 = 0.56 \text{ g CaO} \]

### Step 5: Calculate the Equivalent Mass of Calcium Oxide (CaO)
The equivalent mass of a substance is calculated as:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{n} \]
where \( n \) is the number of equivalents. For CaO, since it can provide 2 moles of hydroxide ions (OH⁻) when dissolved in water, the \( n \) factor is 2:
\[ \text{Equivalent mass of CaO} = \frac{56 \text{ g/mol}}{2} = 28 \text{ g/equiv} \]

### Step 6: Calculate the Number of Equivalents of CaO
The number of equivalents is given by:
\[ \text{Number of equivalents} = \frac{\text{Given mass}}{\text{Equivalent mass}} \]
Using the mass of quicklime produced:
\[ \text{Number of equivalents} = \frac{0.56 \text{ g}}{28 \text{ g/equiv}} = 0.02 \text{ equiv} \]

### Step 7: Calculate the Normality of the Solution
Normality (N) is defined as the number of equivalents per liter of solution:
\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume of solution in liters}} \]
Since the volume of the solution is 1 liter:
\[ \text{Normality} = 0.02 \text{ equiv} / 1 \text{ L} = 0.02 \text{ N} \]

### Step 8: Relate Normality to the Given Expression
According to the problem, the normality of the solution is given as:
\[ 0.01 \times x \]
Setting this equal to the calculated normality:
\[ 0.01 \times x = 0.02 \]

### Step 9: Solve for x
To find the value of \( x \):
\[ x = \frac{0.02}{0.01} = 2 \]

### Final Answer
The value of \( x \) is **2**.

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