Which of the following expression correctly represents the
equivalent conductance at infinite dilution of `Ca_(3)(PO_(4))_(2)` Given `^^_(Ca^(+2))^(@)` and `^^_(PO_(4)^(-3))^(@)` are the equivalent conductance at infinity
dilution of the respective ions :-

Options

A. `^^_(Ca^(+2))^(@) +^^_(PO_(4)^(-3))^(@)`
B. `(^^_(Ca^(+2))^(@) +^^_(PO_(4)^(-3))^(@))xx6`
C. `(1)/(2)^^_(Ca^(+2))^(@) +(1)/(3)^^_(PO_(4)^(-3))^(@)`
D. `3^^_(Ca^(+2))^(@) +2^^_(PO_(4)^(-3))^(@)`

Answer

To find the equivalent conductance at infinite dilution of \( Ca_3(PO_4)_2 \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( Ca_3(PO_4)_2 \) in water can be represented as:
\[ Ca_3(PO_4)_2 \rightarrow 3Ca^{2+} + 2PO_4^{3-} \]

### Step 2: Identify the ions produced
From the dissociation equation, we can see that the compound produces:
- 3 moles of \( Ca^{2+} \) ions
- 2 moles of \( PO_4^{3-} \) ions

### Step 3: Apply Kohlrausch's Law
According to Kohlrausch's Law, the equivalent conductance at infinite dilution (\( \Lambda^0 \)) of an electrolyte is the sum of the equivalent conductance of its constituent ions. Thus, we can express the equivalent conductance of \( Ca_3(PO_4)_2 \) as:
\[ \Lambda^0_{Ca_3(PO_4)_2} = 3\Lambda^0_{Ca^{2+}} + 2\Lambda^0_{PO_4^{3-}} \]

### Step 4: Write the final expression
The final expression for the equivalent conductance at infinite dilution of \( Ca_3(PO_4)_2 \) is:
\[ \Lambda^0_{Ca_3(PO_4)_2} = 3 \cdot \Lambda^0_{Ca^{2+}} + 2 \cdot \Lambda^0_{PO_4^{3-}} \]

### Conclusion
Thus, the expression that correctly represents the equivalent conductance at infinite dilution of \( Ca_3(PO_4)_2 \) is:
\[ \Lambda^0_{Ca_3(PO_4)_2} = 3 \Lambda^0_{Ca^{2+}} + 2 \Lambda^0_{PO_4^{3-}} \]

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