Q.Calculate $\Lambda^0_m$ for $CaCl_2$ and $MgSO_4$ from the data given in Table 3.4.
(The relevant limiting molar conductivities are: $\lambda^0(Ca^{2+}) = 119.0$, $\lambda^0(Cl^-) = 76.3$, $\lambda^0(Mg^{2+}) = 106.0$ and $\lambda^0(SO_4^{2-}) = 160.0\ S\ cm^2\ mol^{-1}$.)

Concept: Molar Conductivity — Kohlrausch’s law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions, each multiplied by its stoichiometric coefficient.

Step 1: For $CaCl_2$

$CaCl_2$ dissociates as $Ca^{2+} + 2Cl^-$.

Using Kohlrausch’s law:

Step 2: Substitute values

Step 3: For $MgSO_4$

$MgSO_4$ dissociates as $Mg^{2+} + SO_4^{2-}$.

The key idea is that at infinite dilution, ions behave completely independently — they don't interact with each other. So the total conductivity of a salt solution is simply the sum of the contributions from each type of ion, each multiplied by how many of that ion appear in the formula unit.

This is Kohlrausch's law of independent migration. It's a powerful shortcut: you don't need to measure every salt directly. Once you know the limiting molar conductivity of a few key ions, you can predict $\Lambda^0_m$ for any salt made from them.

Let's apply it.

1. For $CaCl_2$ One formula unit gives one $Ca^{2+}$ ion and two $Cl^-$ ions. So:

Plug in the numbers:

2. For $MgSO_4$ One formula unit gives one $Mg^{2+}$ and one $SO_4^{2-}$ ion. So:

Substituting:

Method: Kohlrausch’s Law of Independent Migration of Ions

This law states that at infinite dilution, each ion contributes a fixed amount to the total molar conductivity of an electrolyte, independent of the other ion present.

Steps

1. Recall the formula For any electrolyte $A_xB_y$:

where $x$ and $y$ are the number of cations and anions per formula unit.

2. For $CaCl_2$
   • $CaCl_2$ dissociates as: $Ca^{2+} + 2Cl^-$
   • So $x = 1$, $y = 2$
   • Using given values:

3. For $MgSO_4$
   • $MgSO_4$ dissociates as: $Mg^{2+} + SO_4^{2-}$
   • So $x = 1$, $y = 1$
   • Using given values:

Key Concept Check

• Why does this work? At infinite dilution, ions are so far apart that they don’t interact — each ion’s conductivity is purely its own property.
• Units note: All values are in $S\ cm^2\ mol^{-1}$ — always include units in your final answer for exams.

Common Mistakes Students Make with Molar Conductivity (and How to Avoid Them)

Mistake 1: Forgetting to Multiply by Stoichiometric Coefficients

The error:

Students often directly add the given ionic conductivities without considering the number of ions in the formula unit. For example, for $CaCl_2$, they write:

This is wrong because $CaCl_2$ has two chloride ions.

How to avoid:

Always write the dissociation equation first:

Then apply Kohlrausch’s law correctly:

where $\nu$ is the number of ions of each type.

Correct calculation:

Mistake 2: Confusing the Formula for 1:1 vs 2:2 Electrolytes

The error:

For $MgSO_4$, students sometimes incorrectly multiply both ions by 2, thinking "both are divalent so double everything."

How to avoid:

Remember: stoichiometry matters, not just charge. $MgSO_4$ dissociates as:

There is one magnesium ion and one sulfate ion. So:

Key insight: Charge tells you the mobility (given in the table), but the count of ions tells you the multiplier.

Mistake 3: Mixing Up Units or Omitting Them

The error:

Students write numbers without units, or confuse $S\ cm^2\ mol^{-1}$ with $S\ m^2\ mol^{-1}$.

How to avoid:

• Always attach units to your final answer.
• In NCERT/board exams, the standard unit is $S\ cm^2\ mol^{-1}$.
• If conversion is needed: $1\ S\ cm^2\ mol^{-1} = 10^{-4}\ S\ m^2\ mol^{-1}$.

Correct final answers with units:

• $\Lambda^0_m(CaCl_2) = 271.6\ S\ cm^2\ mol^{-1}$
• $\Lambda^0_m(MgSO_4) = 266.0\ S\ cm^2\ mol^{-1}$

Mistake 4: Using the Wrong Table Values

The error:

Students accidentally swap values (e.g., using $\lambda^0(Mg^{2+})$ for $Ca^{2+}$) or misread the table.

How to avoid:

• Label each value as you copy it from the table.
• Double-check: $Ca^{2+} = 119.0$, $Cl^- = 76.3$, $Mg^{2+} = 106.0$, $SO_4^{2-} = 160.0$.
• Cross-check with periodic trends: $Ca^{2+}$ has higher conductivity than $Mg^{2+}$ (larger ion, less hydration), so $119 > 106$ makes sense.

Quick Checklist to Avoid All Mistakes

Final correct answers for reference: