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 is (in m2 mho mol-1)

Question

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 is (in m2 mho mol-1) (A) 118.88 × 10-4 (B) 154.66 × 10-4 (C) 273.54 × 10-4 (D) 196.21 × 10-4

Tardigrade Admin · Accepted Answer

From Kohlrausch's law: Lambdam∞=v+ λ+∞+v- λ-∞ For CaCl 2: Lambdam∞( CaCl 2)=λ Ca 2+∞+2 λ Cl -∞ =118.88 × 10-4+2 × 77.33 × 10-4 =118.88 × 10-4+154.66 × 10-4 =273.54 × 10-4 m 2 mho mol -1

Q. If the molar conductance values of $Ca X^{2 +}$ and $Cl X^{-}$ at infinite dilution are respectively $118.88 \times 10 X^{- 4} m X^{2} mho mol X^{- 1}$ and $77.33 \times 10 X^{- 4} m X^{2} mho mol X^{- 1}$ then that of $CaCl X_{2}$ is (in $m X^{2} mho mol X^{- 1}$ )

$118.88 \times 1 0^{- 4}$

$154.66 \times 1 0^{- 4}$

$273.54 \times 1 0^{- 4}$

$196.21 \times 1 0^{- 4}$

Q. If the molar conductance values of CaX2+ and ClX− at infinite dilution are respectively 118.88×10X−4 mX2 mho molX−1 and 77.33×10X−4 mX2 mho molX−1 then that of CaClX2​ is (in mX2 mho molX−1)

118.88×10−4

154.66×10−4

273.54×10−4

196.21×10−4

From Kohlrausch's law : Λm∞​=v+​λ+∞​+v−​λ−∞​ For CaCl2​: Λm∞​(CaCl2​)=λCa2+∞​+2λCl−∞​ =118.88×10−4+2×77.33×10−4 =118.88×10−4+154.66×10−4 =273.54×10−4m2mhomol−1

Q. If the molar conductance values of $Ca X^{2 +}$ and $Cl X^{-}$ at infinite dilution are respectively $118.88 \times 10 X^{- 4} m X^{2} mho mol X^{- 1}$ and $77.33 \times 10 X^{- 4} m X^{2} mho mol X^{- 1}$ then that of $CaCl X_{2}$ is (in $m X^{2} mho mol X^{- 1}$ )

$118.88 \times 1 0^{- 4}$

$154.66 \times 1 0^{- 4}$

$273.54 \times 1 0^{- 4}$

$196.21 \times 1 0^{- 4}$