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Question 14 Marks
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer
Conductivity $(\kappa)$ : The inverse (opposite) of resistivity (or specific resistance) is called conductivity. It is also called specific conductivity. Its symbol is $\kappa$ and $\kappa=\frac{1}{\rho}$.
Or, The conductivity of the solution at a given concentration is the conductivity of its unit volume placed between two electrodes at unit cross-sectional area spaced at unit distance.
Molar conductivity ($\Lambda_{ m }$) : Molar conductivity is the conductivity of Volume V of the solution in which one mole of electrolyte is dissolved and which is placed between two electrodes of cross-sectional area A, located at a unit distance from each other.
Or, Molar conductivity is the conductivity of that volume of solution of an electrolyte which is placed between the electrodes located at unit distance of the conductivity cell and whose cross-sectional area is sufficient to hold the volume (V) of the solution in which the electrolyte is present. Therefore, the conductivity of ions obtained by dissolving one mole of electrolyte in solution is called molar conductivity.
Both conductivity and molar conductivity changes with change in concentration of electrolyte. Conductivity always decreases when concentration of both strong and weak electrolytes is reduced because on increasing dilution the number of ions carrying the number of ions carrying electric current per unit volume decreases.
Molar conductivity increases as the concentration decreases because the total volume (V) in which one mole of electrolyte is present increases $\left(\Lambda_{ m } \kappa V \right)$ and the increase in volume is greater than decrease in conductivity $(\kappa)$.
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Question 24 Marks
Conductivity of 0.00241 M acetic acid is $7.896 \times 10^{-5} s cm ^{-1}$. Calculate its molar conductivity. If $\Lambda_m{ }^o$ for acetic acid is 390.5 S cm2 mol-1. What is its dissociation constant?
Answer
Molar conductivity $\left(\Lambda_{ m }\right)=\frac{\kappa \times 1000}{\text { Molarity }}$
Conductivity $(\kappa)=7.896 \times 10^{-5} S cm ^{-1}$
Molarity = 0.00241 M
Hence, $\Lambda_{ m }=\frac{7.896 \times 10^{-5} \times 1000}{0.00241}$
$\Lambda_{ m }=32.76$ $S$ $cm ^2$ $mol^{-1}$
Degree of dissociation $\alpha=\frac{\Lambda_{ m }}{\Lambda_{ m }^{\circ}}$
$\Lambda_{ m }^{\circ}=390.5$ $S$ $cm ^2$ $mol^{-1}$
Hence, $\alpha=\frac{32.76}{390.5}=8.4 \times 10^{-2}$
Dissociation constant $\left(K_a\right)=\frac{c \alpha^2}{1-\alpha}$
$\alpha=8.4 \times 10^{-2}$
Hence, $K _{ a }=\frac{0.00241 \times\left(8.4 \times 10^{-2}\right)^2}{1-8.4 \times 10^{-2}}$
$K _{ a }=\frac{0.00241 \times 70.56 \times 10^{-4}}{1-0.084}$
$K _{ a }=\frac{0.1700 \times 10^{-4}}{0.916}$
$K _{ a }=0.1855 \times 10^{-4}$
$K _{ a }=1.85 \times 10^{-5}$
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4 Marks Questions - Chemistry STD 12 Science Questions - Vidyadip