Question

calculate molar conductivity at infinite dilution( ^0m) for CaCl2 and MgCl2 from following data: (scm2mol-1):Ca2+=119.0,
Mg2+=106.0,Cl-=76.3,SO4^2-=160.0​

Answer 1

Answer:

Explanation:

for CaCl2

Ca^2+ + 2Cl-=199+2×76.3

=351.6

forMgSO4

Mg^2+ +SO4^2-=106+160

=266

Answer 2

The molar conductivity at infinite dilution of $CaCl_2\text{ and }MgCl_2$ is $271.6Scm^2$  and $258.6Scm^2$

Explanation:

To calculate the equivalent conductance of $CaCl_2\text{ and }MgCl_2$ at infinite dilution, we will apply Kohlrausch Law.

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,

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

• The ionization equation for calcium chloride follows the equation:

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

The equation for the Kohlrausch law of the given electrolyte follows:

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

We are given:

$\lambda^o_{(Ca^{2+})}=119.0Scm^2mol^{-1}\\\lambda^o_{(Cl^{-})}=76.3Scm^2mol^{-1}$

Putting values in above equation, we get:

$\Lambda^o_{CaCl_2}=[119.0+(2\times 76.3)]\\\\\Lambda^o_{CaCl_2}=271.6Scm^2$

• The ionization equation for magnesium chloride follows the equation:

$MgCl_2\rightarrow Mg^{2+}+2Cl^{-}$

The equation for the Kohlrausch law of the given electrolyte follows:

$\Lambda^o_m\text{ for }MgCl_2=\lambda^o_{(Mg^{2+})}+2\lambda^o_{(Cl^-)}$

We are given:

$\lambda^o_{(Mg^{2+})}=106.0Scm^2mol^{-1}\\\lambda^o_{(Cl^{-})}=76.3Scm^2mol^{-1}$

Putting values in above equation, we get:

$\Lambda^o_{MgCl_2}=[106.0+(2\times 76.3)]\\\\\Lambda^o_{MgCl_2}=258.6Scm^2$

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