If the molar conductance values of Ca2- and Cl- at infinite dilution are respectively 118-88- 10 -4 m2 mho mol-1 and 77-33- 10 -4 m2 mho mol-1- then that of CaCl2 in m2 mho mol-1 is-

Λ _m^∘ (CaCl_2)= λ _m^∘ (Ca^2+)+ 2 λ _m^∘ (Cl^ ) = 118.88× 10^ 4+ 2× (77.33 × 10^ 4) =273.54 × 10^ 4 m^2 mho mol^ 1 ...

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Standard X

Chemistry

Electrolysis and Electrolytes

If the molar ...

Question

If the molar conductance values of $C a^{2 +}$ and $C l^{-}$ at infinite dilution are respectively $118.88 \times 10^{- 4} m^{2} {mho mol}^{- 1}$ and $77.33 \times 10^{- 4} m^{2} {mho mol}^{- 1}$ , then that of $C a C l_{2}$ $(in m^{2} {mho mol}^{- 1})$ is:

118.88 \times 10^{- 4}

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154.66 \times 10^{- 4}

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273.54 \times 10^{- 4}

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196.21 \times 10^{- 4}

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Solution

The correct option is C $273.54 \times 10^{- 4}$
$λ_{m}^{\circ} (C a C l_{2}) = λ_{m}^{\circ} (C a^{2 +}) + 2 λ_{m}^{\circ} (C l^{-}) = 118.88 \times 10^{- 4} + 2 \times (77.33 \times 10^{- 4}) = 273.54 \times 10^{- 4} m^{2} {mho mol}^{- 1}$

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Q. If the molar conductance values of

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If the molar conductance values of Ca2+ and Cl− at infinite dilution are respectively 118.88×10−4 m2 mho mol−1 and 77.33×10−4 m2 mho mol−1, then that of CaCl2 (in m2 mho mol−1) is:

The correct option is C 273.54×10−4λ∘m(CaCl2)=λ∘m(Ca2+)+2λ∘m(Cl−)=118.88×10−4+2×(77.33×10−4)=273.54×10−4 m2 mho mol−1

If the molar conductance values of $C a^{2 +}$ and $C l^{-}$ at infinite dilution are respectively $118.88 \times 10^{- 4} m^{2} {mho mol}^{- 1}$ and $77.33 \times 10^{- 4} m^{2} {mho mol}^{- 1}$ , then that of $C a C l_{2}$ $(in m^{2} {mho mol}^{- 1})$ is:

The correct option is C $273.54 \times 10^{- 4}$
$λ_{m}^{\circ} (C a C l_{2}) = λ_{m}^{\circ} (C a^{2 +}) + 2 λ_{m}^{\circ} (C l^{-}) = 118.88 \times 10^{- 4} + 2 \times (77.33 \times 10^{- 4}) = 273.54 \times 10^{- 4} m^{2} {mho mol}^{- 1}$