Given below are two statements:
Statement I: Mohr's salt is composed of only three types of ions—ferrous, ammonium, and sulphate.
Statement II: If the molar conductance at infinite dilution of ferrous, ammonium, and sulphate ions are $ x_1 $, $ x_2 $, and $ x_3 $ $ \text{S cm}^2 \, \text{mol}^{-1} $, respectively, then the molar conductance for Mohr's salt solution at infinite dilution would be given by $ x_1 + x_2 + 2x_3 $.
Step 1: Analyzing Statement I
Mohr's salt is composed of three ions: ferrous \( \text{Fe}^{2+} \), ammonium \( \text{NH}_4^+ \), and sulphate \( \text{SO}_4^{2-} \).
This statement is true because Mohr's salt (also known as ammonium ferrous sulphate) contains only these three types of ions.
Thus, Statement I is correct.
Step 2: Analyzing Statement II
The molar conductance of a salt at infinite dilution is the sum of the molar conductances of the ions it dissociates into. For Mohr's salt, the ions involved are:
Ferrous ion \( \text{Fe}^{2+} \), with a molar conductance of \( x_1 \), Ammonium ion \( \text{NH}_4^+ \), with a molar conductance of \( x_2 \), Sulphate ion \( \text{SO}_4^{2-} \), with a molar conductance of \( x_3 \). At infinite dilution, the total molar conductance \( \lambda_{\infty} \) of Mohr's salt should be the sum of the conductances of these ions. However, Statement II suggests the wrong coefficient for the sulphate ion. The sulphate ion \( \text{SO}_4^{2-} \) is a monoatomic ion and should contribute \( x_3 \) to the total conductance, not \( 2x_3 \).
Thus, the correct expression for the molar conductance should be: \[ \lambda_{\infty} = x_1 + x_2 + x_3 \]
Therefore, Statement II is incorrect because it incorrectly doubles the contribution of the sulphate ion.
Match List-I with List-II: List-I
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)