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 problem of determining the molar conductivity of the compound \( K_2[PtCl_4] \) in aqueous solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dissociation of the Compound:**
The compound \( K_2[PtCl_4] \) dissociates in water into its constituent ions. The dissociation can be represented as:
\[ K_2[PtCl_4] \rightarrow 2K^+ + [PtCl_4]^{2-} \]
Here, \( K_2[PtCl_4] \) produces 2 potassium ions (\( K^+ \)) and one complex ion (\( [PtCl_4]^{2-} \)).

2. **Count the Total Number of Ions Produced:**
From the dissociation, we see that:
- 2 \( K^+ \) ions
- 1 \( [PtCl_4]^{2-} \) ion
Therefore, the total number of ions produced is:
\[ 2 + 1 = 3 \text{ ions} \]

3. **Understanding Molar Conductivity:**
Molar conductivity (\( \Lambda_m \)) is directly related to the number of ions produced in solution and their charges. More ions typically lead to higher molar conductivity.

4. **Comparing with Other Compounds:**
Now, we need to compare the molar conductivity of \( K_2[PtCl_4] \) with the given options:
- **KNO3:** Dissociates into \( K^+ \) and \( NO_3^- \) giving 2 ions.
- **CCl4:** Does not dissociate in water (non-polar).
- **MgSO4:** Dissociates into \( Mg^{2+} \) and \( SO_4^{2-} \) giving 2 ions.
- **Na2SO4:** Dissociates into \( 2Na^+ \) and \( SO_4^{2-} \) giving 3 ions.

5. **Determine the Closest Molar Conductivity:**
Since \( K_2[PtCl_4] \) produces 3 ions, we look for a compound that also produces 3 ions. The only compound that fits this criterion is \( Na_2SO_4 \), which also produces 3 ions:
\[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \]

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