If the molar conductance values of Ca -2 and Cl -at infinite dilution are respectively 1118-88 - 10-4 m 2 ohm -1 mol -1 and 77-33 - 10-4 m 2 ohm -1 mol -1 then that of CaCl 2 is in m 2 ohm -1 mol -1 A- 154-66 - 10-4B- 118-88 - 10-4C- 273-54 - 10-4D- 196-21 - 10-4

The correct option is C 273.54×10 4 Λ _m^0CaCl_2=λ^0Ca^+2+2λ^0Cl^ =(118.88× 10^ 4)+2(77.33× 10^ 4) =273.54× 10^ 4 m^2 ohm^ 1 mol^ 1 ...

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Standard XII

Chemistry

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Question

If the molar conductance values of $C a^{+ 2}$ and $C l^{-}$ at infinite dilution are respectively $1118.88 \times 10^{- 4} m^{2} o h m^{- 1} m o l^{- 1} and 77.33 \times 10^{- 4} m^{2} o h m^{- 1} m o l^{- 1}$ then that of $C a C l_{2}$ is (in $m^{2} o h m^{- 1} m o l^{- 1}$ )

118.88×10^-4

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154.66×10^-4

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273.54×10^-4

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196.21×10^-4

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Solution

The correct option is C
273.54×10^-4

$Λ_{m}^{0} C a C l_{2} = λ^{0} C a^{+ 2} + 2 λ^{0} C l^{-} = (118.88 \times 10^{- 4}) + 2 (77.33 \times 10^{- 4}) = 273.54 \times 10^{- 4} m^{2} o h m^{- 1} m o l^{- 1}$

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If the molar conductance values of Ca+2 and Cl− at infinite dilution are respectively 1118.88×10−4m2 ohm−1mol−1 and 77.33×10−4m2 ohm−1mol−1 then that of CaCl2 is (in m2 ohm−1 mol−1 )

The correct option is C 273.54×10-4 Λ0mCaCl2=λ0Ca+2+2λ0Cl−=(118.88×10−4)+2(77.33×10−4)=273.54×10−4 m2ohm−1 mol−1

If the molar conductance values of $C a^{+ 2}$ and $C l^{-}$ at infinite dilution are respectively $1118.88 \times 10^{- 4} m^{2} o h m^{- 1} m o l^{- 1} and 77.33 \times 10^{- 4} m^{2} o h m^{- 1} m o l^{- 1}$ then that of $C a C l_{2}$ is (in $m^{2} o h m^{- 1} m o l^{- 1}$ )

The correct option is C
273.54×10^-4

$Λ_{m}^{0} C a C l_{2} = λ^{0} C a^{+ 2} + 2 λ^{0} C l^{-} = (118.88 \times 10^{- 4}) + 2 (77.33 \times 10^{- 4}) = 273.54 \times 10^{- 4} m^{2} o h m^{- 1} m o l^{- 1}$