Question

42. If limiting molar conductances of Ca+2 and Clions
are 119 and 76.3 Scm? mol-respectively. The
limiting molar conductivity of CaCl, is :-
(1) 347 Scm? mol-1
(2) 271.6 Scmmol-1
(3) 42.7 Scrnmot
(4) 195.3 Scm? mol1
​

Answer 1

119 + (76.3×2) = 271.6

Answer 2

The molar conductivity at infinite dilution for $CaCl_2$ is $271.6Scm^2$

Explanation:

Kohlrausch law of limiting molar conductivity is defined as the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte.

Mathematically,

$\Lambda^o_m\text{ for }A_xB_y=x\lambda^o_++y\lambda^o_-$

Calculating the molar conductivity at infinite dilution for $CaCl_2$ :

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

The equation used to calculate the limiting molar conductivity of $CaCl_2$ follows:

$\Lambda^o_{CaCl_2}=\lambda^o_{(Ca^{2+})}+2\lambda^o_{(Cl^-)}$

We are given:

$\lambda^o_{(Ca^{2+})}=119Scm^2mol^{-1}\\\\\lambda^o_{(Cl^-)}=76.3Scm^2mol^{-1}$

Putting values in above equation, we get:

$\Lambda^o_{CaCl_2}=[119+(2\times 76.3)]\\\\\Lambda^o_{CaCl_2}=271.6Scm^2$

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