Question

The molar conductivities of infinite dilution for sodium iodide, sodium acetate and aluminium acetate are 12.69, 9.10 and 24.52 s cm2 mol−1 respectively at 25 °c. what is the molar conductivity of ali3 at infinite dilution?

Answer 1

According to Kohlrausch's law , molar conductivities of dilute solution for sodium iodide is represented by
$\Lambda^0_{NaI}=\Lambda^0_{Na^+}+\Lambda^0_{I^-}=12.69$----(1)
Similarly for sodium acetate is represented by,$\Lambda^0_{CH_3COONa}=\Lambda^0_{Na^+}+\Lambda^0_{CH_3COO^-}=9.10$-----(2)
and for aluminate acetate is represented by, $\Lambda^0_{(CH_3COO)_3Na}=\Lambda^0_{Al^{3+}}+3\Lambda^0_{CH_3COO^-}=24.52$-------(3)

Now, we have to find out, $\Lambda^0_{AlI_3}=\Lambda^0_{Al^{3+}}+3\Lambda^0_{I^-}$

Do 3 × equation (1) + equation (3) - 3 × equation (2)
We get , $\Lambda^0_{AlI_3}=\Lambda^0_{Al^{3+}}+3\Lambda^0_{I^-}$
= 3 × 12.69 + 24.52 - 3 × 9.1
= 35.29 cm²mol⁻¹

Hence, answer is 35.29 cm²mol⁻¹

Answer 2

The value of molar conductivity of Al3+ is 35.29 cm^2 / mol.

Explanation:

The equations of molar conductivities for sodium iodide, sodium acetate, and aluminum acetate are 12.69, 9.10, and 24.52 s cm2 mol−.

Now

• [ λ Na+ + λI- = 12.69 cm^2 / mol  ]  x 3  ------(1)
• [ λ Na+ + CH3COO- = 9.10 cm^2 / mol  ]   x  3  -------(2)
• [ λ Al3+  + 3 λ CH3COO- = 24.52 cm^2 / mol ]    ----(3)  

Now multiply all the equation by 3.

Now add equations 1 + 3 and then subtract from equation 2.

λ Al3+   + 3 λ CH3COO- + 3 V Na+  + 3 λ I- - 3 λ Na+ - 3 λ CH3COO-

λAl3+ + 3 λ I- = 35.29