Question:medium

\(3\times 10^{22}\) molecules of \(Na_2CO_3\) \((\text{molecular weight}=106)\) are present in \(500\) ml of solution. The normality of the solution formed is \((N_A=6\times 10^{23}\,mol^{-1})\)

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For \(Na_2CO_3\), normality is twice the molarity because its \(n\)-factor is \(2\).
  • \(0.1\,N\)
  • \(0.2\,N\)
  • \(0.4\,N\)
  • \(0.05\,N\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This problem requires calculating the normality of a solution. Normality is defined as the number of gram equivalents of solute per liter of solution. The calculation involves finding the number of moles, then the molarity, and finally the normality using the n-factor of the solute.
Step 2: Key Formula or Approach:
1. Calculate moles (n): $n = \frac{\text{Number of molecules}}{\text{Avogadro's number (N}_A\text{)}}$. 2. Calculate Molarity (M): $M = \frac{\text{moles (n)}}{\text{Volume in Liters (V)}}$. 3. Determine n-factor: For a salt, the n-factor is the total magnitude of the positive or negative charge of the ions produced upon dissociation. 4. Calculate Normality (N): $N = M \times \text{n-factor}$.
Step 3: Detailed Explanation:
Given: - Number of molecules = $3 \times 10^{22}$. - Avogadro's number, N$_A = 6 \times 10^{23}$ mol$^{-1}$. - Volume of solution, V = 500 ml = 0.5 L. - Solute is sodium carbonate, Na$_2$CO$_3$. 1. Calculate moles of Na$_2$CO$_3$: \[ n = \frac{3 \times 10^{22}}{6 \times 10^{23}} = \frac{3}{60} = \frac{1}{20} = 0.05 \text{ moles} \] 2. Calculate the Molarity of the solution: \[ M = \frac{n}{V} = \frac{0.05 \text{ mol}}{0.5 \text{ L}} = 0.1 \text{ M} \] 3. Determine the n-factor for Na$_2$CO$_3$: Sodium carbonate is a salt that dissociates as: 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 $-2$. Therefore, the n-factor (or valence factor) is 2. 4. Calculate the Normality of the solution: \[ N = M \times \text{n-factor} \] \[ N = 0.1 \times 2 = 0.2 \text{ N} \] Step 4: Final Answer:
The normality of the solution is 0.2 N. Therefore, option (B) is correct.
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