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Molar ionic conductance of
${Ca}^{+ 2} is ' x^{'} {Sm}^{2}$ ${mole}^{- 1} and that of {PO}_{4}^{- 3} is ' y' {Sm}^{2} {mole}^{- 1}$
Equivalent conductance of calcium phosphate $is.... Sm {Smeq}^{2}$

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x + y

(3x + 2y)

$\frac{3 x + 2 y}{6}$

6(3x + 2y)

answer is D.

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Detailed Solution

$λ_{{Ca}_{3} {({PO}_{4})}_{2}}^{M} = 3 λ_{{Ca}^{+ 2}}^{M} + 2 λ_{{PO}_{4}^{- 3}}^{M} = 3 x + 2 y$

$λ_{{Ca}_{3} {({PO}_{4})}_{2}}^{eq} = \frac{λ_{{Ca}_{3} {({PO}_{4})}_{2}}^{M}}{6} = \frac{3 x + 2 y}{6}$

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