Question:medium

Calculate the number of \(\text{Cl}^-\) ions in 222 g unhydrous calcium chloride?
(At. mass \(\text{Ca} = 40, \text{Cl} = 35.5\) )

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The direct method is: \[ \text{Number of ions} = \text{moles of compound} \times \text{ions per formula unit} \times N_A \]
Updated On: May 14, 2026
  • \(\text{N}_{\text{A}}\)
  • \(2 \text{N}_{\text{A}}\)
  • \(3 \text{N}_{\text{A}}\)
  • \(4 \text{N}_{\text{A}}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
To find the number of ions in a given mass of a compound, we first need to determine the number of moles of the compound.
Then, we use the chemical formula to find the relationship between moles of the compound and moles of the specific ion.
Finally, we convert moles to the actual number of particles using Avogadro's number (\(\text{N}_{\text{A}}\)).
Step 2: Key Formula or Approach:
First, calculate the moles of \(\text{CaCl}_2\) using \(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\).
Then, use the stoichiometry of the salt to find the moles of \(\text{Cl}^-\) and multiply by Avogadro's number (\(\text{N}_{\text{A}}\)).
Step 3: Detailed Explanation:
First, calculate the molar mass of unhydrous calcium chloride (\(\text{CaCl}_2\)).
\[ \text{Molar Mass of } \text{CaCl}_2 = \text{Atomic mass of Ca} + 2 \times \text{Atomic mass of Cl} \] \[ \text{Molar Mass} = 40 + 2(35.5) = 40 + 71 = 111 \text{ g/mol} \] Next, calculate the number of moles of \(\text{CaCl}_2\) in \(222 \text{ g}\).
\[ \text{Moles of } \text{CaCl}_2 = \frac{222 \text{ g}}{111 \text{ g/mol}} = 2 \text{ moles} \] From the chemical formula \(\text{CaCl}_2\), we can see that one molecule of \(\text{CaCl}_2\) dissociates to give one \(\text{Ca}^{2+}\) ion and two \(\text{Cl}^-\) ions.
Therefore, 1 mole of \(\text{CaCl}_2\) yields 2 moles of \(\text{Cl}^-\) ions.
So, \(2 \text{ moles}\) of \(\text{CaCl}_2\) will yield:
\[ \text{Moles of } \text{Cl}^- \text{ ions} = 2 \text{ moles of } \text{CaCl}_2 \times 2 = 4 \text{ moles} \] Finally, calculate the total number of \(\text{Cl}^-\) ions.
\[ \text{Number of } \text{Cl}^- \text{ ions} = 4 \times \text{N}_{\text{A}} \] Step 4: Final Answer:
The total number of \(\text{Cl}^-\) ions is \(4 \text{N}_{\text{A}}\).
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