Step 1: Understanding the Dissociation of Salt MX\(_3\)
The salt MX\(_3\) dissociates in water as follows: \[ \text{MX}_3 \rightarrow \text{M}^{3+} + 3\text{X}^- \] For each mole of MX\(_3\), it dissociates to give one mole of M\(^{3+}\) and three moles of X\(^-\).
Step 2: van't Hoff Factor (i)
The van't Hoff factor \(i\) is given as \(i = 2\). The van't Hoff factor represents the total number of particles in solution per formula unit of solute. In this case, for MX\(_3\), the dissociation would produce 4 particles (1 M\(^{3+}\) and 3 X\(^-\)) per formula unit of MX\(_3\).
However, since \(i = 2\), this suggests that the dissociation is not complete, and the actual number of particles formed is only double the initial number of formula units.
Step 3: Using the Formula for Percentage Dissociation
The formula for the percentage dissociation (\(\alpha\)) is given by: \[ i = 1 + \alpha (n - 1) \] Where:
\(i\) is the van't Hoff factor (2 in this case), \(\alpha\) is the degree of dissociation, \(n\) is the number of ions produced per formula unit of solute (which is 4 for MX\(_3\)).
\[ 2 = 1 + \alpha (4 - 1) \] \[ 2 = 1 + 3\alpha \] \[ 3\alpha = 1 \] \[ \alpha = \frac{1}{3} \]
Step 4: Calculating the Percentage Dissociation
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 = \frac{1}{3} \times 100 = 33.33% \] To the nearest integer, the percentage dissociation is 33%.
Given below are two statements:
Statement (I): Molal depression constant $ k_f $ is given by $ \frac{M_1 R T_f}{\Delta S_{\text{fus}}} $, where symbols have their usual meaning.
Statement (II): $ k_f $ for benzene is less than the $ k_f $ for water.
In light of the above statements, choose the most appropriate answer from the options given below:
In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)
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