`1 g` of calcium was burnt in excess of `O_(2)` and the oxide was dissolved in water to make up `1 L` solution. Calculate the normality of alkaline soluiton.

Options

A. 0.04
B. 0.4
C. 0.05
D. 0.5

Answer

To calculate the normality of the alkaline solution formed when 1 g of calcium is burnt in excess oxygen, we can follow these steps:

### Step 1: Determine the moles of calcium
First, we need to calculate the number of moles of calcium (Ca) present in 1 g.

\[ \text{Molar mass of Ca} = 40 \, \text{g/mol} \]

\[ \text{Moles of Ca} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \, \text{g}}{40 \, \text{g/mol}} = 0.025 \, \text{mol} \]

### Step 2: Write the reaction for calcium burning
Calcium reacts with oxygen to form calcium oxide (CaO):

\[ 2 \, \text{Ca} + \text{O}_2 \rightarrow 2 \, \text{CaO} \]

From this reaction, we see that 1 mole of calcium produces 1 mole of calcium oxide.

### Step 3: Determine the moles of calcium oxide produced
Since we have 0.025 moles of calcium, the moles of calcium oxide produced will also be:

\[ \text{Moles of CaO} = 0.025 \, \text{mol} \]

### Step 4: Write the reaction for calcium oxide with water
Calcium oxide reacts with water to form calcium hydroxide (Ca(OH)₂):

\[ \text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 \]

From this reaction, we see that 1 mole of CaO produces 1 mole of Ca(OH)₂.

### Step 5: Determine the moles of calcium hydroxide produced
Thus, the moles of calcium hydroxide produced will also be:

\[ \text{Moles of Ca(OH)}_2 = 0.025 \, \text{mol} \]

### Step 6: Calculate the molarity of the solution
The solution is made up to 1 L (1 liter) of calcium hydroxide. Therefore, the molarity (M) of the solution is:

\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.025 \, \text{mol}}{1 \, \text{L}} = 0.025 \, \text{M} \]

### Step 7: Determine the normality of the solution
Calcium hydroxide (Ca(OH)₂) is a diacidic base, meaning it can donate 2 hydroxide ions (OH⁻) per formula unit. Therefore, the n-factor for Ca(OH)₂ is 2.

\[ \text{Normality} = \text{n-factor} \times \text{Molarity} = 2 \times 0.025 \, \text{M} = 0.05 \, \text{N} \]

### Final Answer
The normality of the alkaline solution is **0.05 N**.

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