Question

3 x 10\(^{22}\) molecules of Na\(_2\)CO\(_3\) (molecular weight = 106) present in 500 ml of solution. The normality of the solution formed is (N = 6 x 10\(^{23}\) mol\(^{-1}\))

• A. 0.1 N
• B. 0.2 N
• C. 0.4 N
• D. 0.05 N

Hint:

The n-factor is crucial for converting between molarity and normality. For acids, it's the number of H⁺ ions; for bases, the number of OH⁻ ions; and for salts, it's the total charge on the cations (or anions).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the normality of a sodium carbonate solution, given the number of molecules, volume, and molecular weight.

Step 2: Key Formula or Approach:
1. Calculate the number of moles of Na\(_2\)CO\(_3\).
2. Calculate the Molarity (M) of the solution.
3. Calculate the Normality (N) using the formula \(N = M \times \text{n-factor}\).

Step 3: Detailed Explanation:

1. Calculate moles of Na\(_2\)CO\(_3\):
The number of moles is the number of molecules divided by Avogadro's number (\(N_A\)).
\[ \text{moles} = \frac{\text{Number of molecules}}{N_A} = \frac{3 \times 10^{22}}{6 \times 10^{23}} = \frac{3}{60} = \frac{1}{20} = 0.05 \text{ mol} \]

2. Calculate Molarity (M):
Molarity is moles of solute per liter of solution. The volume is 500 ml = 0.5 L.
\[ M = \frac{\text{moles}}{\text{Volume (L)}} = \frac{0.05 \text{ mol}}{0.5 \text{ L}} = 0.1 \text{ M} \]

3. Calculate Normality (N):
Normality is Molarity times the n-factor. For a salt like Na\(_2\)CO\(_3\), the n-factor is the total positive (or negative) charge of the ions it dissociates into.
Na\(_2\)CO\(_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-}\)
The total positive charge is \(2 \times (+1) = 2\). The total negative charge is \(-2\). So, the n-factor is 2.
\[ N = M \times \text{n-factor} = 0.1 \text{ M} \times 2 = 0.2 \text{ N} \]

Step 4: Final Answer:
The normality of the solution is 0.2 N.