Question

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

• A. \(0.1\,N\)
• B. \(0.2\,N\)
• C. \(0.4\,N\)
• D. \(0.05\,N\)

Hint:

For \(Na_2CO_3\), normality is twice the molarity because its \(n\)-factor is \(2\).

Answer 1

The Correct Option is B

Step 1: Number of moles is calculated by: \[ \text{Moles}=\frac{\text{Number of molecules}}{N_A} \]

Step 2: Given: \[ \text{Number of molecules}=3\times 10^{22} \] \[ N_A=6\times 10^{23} \]

Step 3: Therefore: \[ \text{Moles}=\frac{3\times 10^{22}}{6\times 10^{23}} \] \[ =\frac{3}{6}\times 10^{-1} \] \[ =0.05 \]

Step 4: Volume of solution: \[ 500\,ml=0.5\,L \]

Step 5: Molarity: \[ M=\frac{0.05}{0.5} \] \[ M=0.1\,M \]

Step 6: For \(Na_2CO_3\), \(n\)-factor is \(2\), because carbonate ion can accept two hydrogen ions. \[ N=M\times n\text{-factor} \] \[ N=0.1\times 2 \] \[ N=0.2\,N \] \[ \boxed{0.2\,N} \]