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Question Bank [2022] — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryQuestion Bank [2022]4 Marks
Question
Calculate molar conductivities at zero concentration for $CaCl_2$​​​​​​​ and $NaCl$.Given: molar ionic conductivities of $Ca^{2+}, Cl^–, Na^+​​​​​​​$​​​​​​​ ions are respectively, $104, 76.4, 50.1 \Omega –1 cm^{–2} mol^{–1}$​​​​​​​
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Answer

Given:
$ \lambda_{ Ca ^{2+}}^0=104 \Omega^{-1} cm ^2 mol ^{-1},$
$\lambda_{ Cl ^{-}}^0=76.4 \Omega^{-1} cm ^2 mol ^{-1},$
$\lambda_{ Na ^{+}}^0=50.1 \Omega^{-1} cm ^2 mol ^{-1}, $
To find: Molar conductivity at zero concentration for $CaCl _2$ and $NaCl$ : $\wedge_0\left( CaCl _2\right)$ and $\wedge_0( NaCl )$
Formula: $\wedge_0= n _{+} \lambda_{+}^0+ n _{-} \lambda_{-}^0$
Calculation: According to Kohlrausch law,
$ \wedge_0\left( CaCl _2\right)=\lambda_{ Ca ^{2+}}^0+2 \lambda_{ Cl ^{-}}^0$
$=104 \Omega^{-1} cm ^2 mol ^{-1}+2 \times 76.4 \Omega^{-1} cm ^2 mol ^{-1}$
$=256.8 \Omega^{-1} cm ^2 mol ^{-1}$
$\wedge_0( NaCl )=\lambda_{ Na ^{+}}^0+\lambda_{ Cl ^{-}}^0$
$=50.1 \Omega^{-1} cm ^2 mol ^{-1}+76.4 \Omega^{-1} cm ^2 mol ^{-1}$
$=126.4 \Omega^{-1} cm ^2 mol ^{-1}$
Molar conductivity at zero concentration of $CaCl _2$ is $2 5 6 . 8 \Omega^{-1} cm ^2 mol ^{-1}$
Molar conductivity at zero concentration of $NaCl$ is $1 2 6 . 4 \Omega^{-1} cm ^2 mol ^{-1}$.

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