The ratio of `(^^m)/(^^eq)` for `Ca_(3) (po_4)_2` will be equal to :

Explanation

To find the ratio of molar conductivity (\( \Lambda_m \)) to equivalent conductivity (\( \Lambda_{eq} \)) for calcium phosphate (\( Ca_3(PO_4)_2 \)), we can follow these steps:

### Step 1: Understand Molar and Equivalent Conductivity
Molar conductivity (\( \Lambda_m \)) is defined as the conductivity of a solution divided by its molarity, while equivalent conductivity (\( \Lambda_{eq} \)) is defined as the conductivity divided by its normality. The formulas are:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
\[
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