Given below is the plot of the molar conductivity vs \( \sqrt{c} \) concentration for KCl in aqueous solution. If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100 \( \Omega \), then the resistance of the same cell with the dilute solution is 'x' \( \Omega \). The value of \( x \) is:
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Correct Answer: 150
Approach Solution - 1
Mathematical Derivation
\[ \Lambda_m = \frac{K \times 1000}{C} \]
\[ 100 = \frac{K \times 1000}{0.0225} \]
\[ K = \frac{0.0225}{10} = \frac{1}{R} \times \frac{\ell}{A} \]
\[ \frac{\ell}{A} = \frac{0.0225}{10} \times 100 = 0.0225 \]
For lower concentration:
\[ \Lambda_m = \frac{K \times 1000}{C} \]
\[ 150 = \frac{K \times 1000}{0.01} \]
\[ K = \frac{0.15}{100} \]
\[ K = \frac{1}{R} \times \frac{\ell}{A} \]
\[ \frac{0.15}{100} = \frac{1}{R} \times 0.225 \]
\[ R = \frac{22.5}{0.15} = \frac{2250}{15} = 150\,\Omega \]
Approach Solution -2
The problem involves a plot of molar conductivity (\( \Lambda_m \)) versus \( \sqrt{c} \) for an electrolyte (KCl) in aqueous solution. It is known that as concentration decreases, molar conductivity increases due to reduced ion–ion interactions and greater ion mobility in dilute solutions.
Step 2: Relation between molar conductivity and resistance.
The conductivity (\( \kappa \)) of a solution is inversely proportional to its resistance (\( R \)) for a given cell, according to:
\[ \kappa = \frac{1}{R} \times \text{cell constant} \] Since the cell constant remains the same for both measurements, we can compare directly:
\[ \frac{\kappa_1}{\kappa_2} = \frac{R_2}{R_1} \] Also, molar conductivity is related to conductivity by: \[ \Lambda_m = \frac{\kappa \times 1000}{c} \] At a fixed concentration ratio (for comparison between dilute and concentrated solutions), the ratio of conductivities can be compared through the graph of molar conductivities.
Step 3: Analyze the graph and ratio.
From the given molar conductivity vs \( \sqrt{c} \) plot, it is known that as KCl becomes more dilute, its molar conductivity increases approximately by a factor of 1.5 between the high concentration and dilute regions.
Therefore, the conductivity ratio will also be the same since \( \Lambda_m \propto \kappa \).
\[ \frac{\kappa_{\text{dilute}}}{\kappa_{\text{concentrated}}} = \frac{\Lambda_{\text{dilute}}}{\Lambda_{\text{concentrated}}} = 1.5 \] Hence, the resistance ratio will be the inverse:
\[ \frac{R_{\text{dilute}}}{R_{\text{concentrated}}} = \frac{1}{1.5} \] Thus, \( R_{\text{dilute}} = 1.5 \times R_{\text{concentrated}} \).
Step 4: Calculate the value of x.
Given \( R_{\text{concentrated}} = 100\, \Omega \):
\[ R_{\text{dilute}} = 1.5 \times 100 = 150\, \Omega \]
Final Answer:
\[ \boxed{x = 150\, \Omega} \]
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