Question

The number of \( \text{Cl}^- \) ions in \(100 \, \text{mL}\) of \(0.001 \, \text{M}\) NaCl solution is:

• A. \(6.022 \times 10^{23}\)
• B. \(6.022 \times 10^{20}\)
• C. \(6.022 \times 10^{19}\)
• D. \(6.022 \times 10^{21}\)

Hint:

Always convert mL to L before using molarity formula!

Answer 1

The Correct Option is C

Concept:
•Molarity \(M = \frac{\text{moles}}{\text{litre}}\)
•Number of particles = moles \( \times N_A\)

Step 1: Convert volume: \[ 100 \, \text{mL} = 0.1 \, \text{L} \]

Step 2: Moles of NaCl: \[ \text{moles} = M \times V = 0.001 \times 0.1 = 10^{-4} \]

Step 3: NaCl dissociates completely: \[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \] \[ \Rightarrow \text{moles of } \text{Cl}^- = 10^{-4} \]

Step 4: \[ \text{Number of ions} = 10^{-4} \times 6.022 \times 10^{23} = 6.022 \times 10^{19} \]