The compound `K_2[PtCl_4]` would have a molar conductivity in aqueous solution most closely approaching that of

Options

A. `KNO_3`
B. `"CCl"_4`
C. `MgSO_4`
D. `Na_2SO_4`

Answer

To solve the question regarding the molar conductivity of the compound \( K_2[PtCl_4] \) in aqueous solution, we can follow these steps:

### Step 1: Understand the dissociation of \( K_2[PtCl_4] \)

When \( K_2[PtCl_4] \) is dissolved in water, it dissociates into its constituent ions. The dissociation can be represented as follows:

\[ K_2[PtCl_4] \rightarrow 2K^+ + [PtCl_4]^{2-} \]

### Step 2: Count the number of ions produced

From the dissociation, we can see that \( K_2[PtCl_4] \) produces:
- 2 potassium ions (\( 2K^+ \))
- 1 complex ion (\( [PtCl_4]^{2-} \))

Thus, the total number of ions produced is:

\[ 2 + 1 = 3 \text{ ions} \]

### Step 3: Identify the molar conductivity of other compounds

Next, we need to find a compound that also produces 3 ions in solution. Let's analyze the options:

1. **\( KNO_3 \)**:
\[ KNO_3 \rightarrow K^+ + NO_3^- \]
Total ions = 2

2. **\( CCl_4 \)**:
- This is a non-electrolyte and does not dissociate.
Total ions = 0

3. **\( MgSO_4 \)**:
\[ MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-} \]
Total ions = 2

4. **\( Na_2SO_4 \)**:
\[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \]
Total ions = 3

### Step 4: Conclusion

From the analysis, we find that \( Na_2SO_4 \) also produces 3 ions in solution, which is the same as \( K_2[PtCl_4] \). Therefore, the molar conductivity of \( K_2[PtCl_4] \) would most closely approach that of \( Na_2SO_4 \).

### Final Answer

The compound \( K_2[PtCl_4] \) would have a molar conductivity in aqueous solution most closely approaching that of \( Na_2SO_4 \).

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