Question

calculate molar conductivity at infinite dilution for CaCl2 MgSO4
from the data given
Ca^2+=199 Simon cm-
Cl- = 76.3
Mg2+ =106
SO4= 160
​

Answer 1

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 for $CaCl_2$ is $351.6Scm^2$ and that of $MgSO_4$ is $266Scm^2$

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.

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

• 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:

$\lambda^o_{(Ca^{2+})}=199Scm^2mol^{-1}\\\\\lambda^o_{(Cl^-)}=76.3Scm^2mol^{-1}$

Putting values in above equation, we get:

$\Lambda^o_{CaCl_2}=[199+(2\times 76.3)]\\\\\Lambda^o_{CaCl_2}=351.6Scm^2$

• Calculating the molar conductivity at infinite dilution for $MgSO_4$ :

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

The equation used to calculate the limiting molar conductivity of $MgSO_4$ follows:

$\Lambda^o_{MgSO_4}=\lambda^o_{(Mg^{2+})}+\lambda^o_{(SO_4^{2-})}$

We are given:

$\lambda^o_{(Mg^{2+})}=106Scm^2mol^{-1}\\\\\lambda^o_{(SO_4^{2-})}=160Scm^2mol^{-1}$

Putting values in above equation, we get:

$\Lambda^o_{MgSO_4}=[106+160]\\\\\Lambda^o_{MgSO_4}=266Scm^2$

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