Electrical conductors and Ostwald’s dilution law Questions in English

(B) Electrical conductivity in an aqueous solution depends on the concentration of ions present.
$HCl$ is a strong acid that undergoes complete dissociation in water: $HCl \rightarrow H^+ + Cl^-$.
$CH_3COOH$ and $HCOOH$ are weak acids that undergo partial dissociation.
$C_6H_{12}O_6$ (glucose) is a non-electrolyte and does not produce ions.
Since $HCl$ provides the highest concentration of ions,it is the best electrical conductor.

(N/A) Ostwald's dilution law states that for a weak electrolyte,the degree of dissociation $(\alpha)$ is inversely proportional to the square root of its concentration $(C)$.
Mathematically,it is expressed as: $\alpha = \sqrt{\frac{K_a}{C}}$,where $K_a$ is the dissociation constant of the weak acid.
As the dilution increases (concentration $C$ decreases),the degree of dissociation $\alpha$ increases,approaching unity at infinite dilution.

(B) For a weak acid,the degree of ionization $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.
Since $K_a$ is constant at a constant temperature,$\alpha \propto \frac{1}{\sqrt{C}}$.
Therefore,$\frac{\alpha_2}{\alpha_1} = \sqrt{\frac{C_1}{C_2}}$.
Given $\alpha_1 = 1 \%$,$C_1 = 0.2 \ M$,and $C_2 = 0.02 \ M$.
Substituting the values: $\frac{\alpha_2}{1 \%} = \sqrt{\frac{0.2}{0.02}} = \sqrt{10}$.
Thus,$\alpha_2 = \sqrt{10} \%$.

(B) For a weak acid,the degree of ionisation $(\alpha)$ is given by $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{K_a \times V}$,where $V$ is the dilution (volume per mole of solute).
Since $K_a$ is constant,$\alpha \propto \sqrt{V}$.
Let the initial volume be $V_1 = 300 \ mL$ and the final volume be $V_2$.
Given that the degree of ionisation doubles,$\alpha_2 = 2\alpha_1$.
Therefore,$\frac{\alpha_1}{\alpha_2} = \sqrt{\frac{V_1}{V_2}} \Rightarrow \frac{1}{2} = \sqrt{\frac{300}{V_2}}$.
Squaring both sides,$\frac{1}{4} = \frac{300}{V_2}$,which gives $V_2 = 1200 \ mL$.
The volume of water to be added is $V_2 - V_1 = 1200 \ mL - 300 \ mL = 900 \ mL$.

(B) For a weak monobasic acid $(HA)$,the dissociation is given by: $HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$.
The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1-\alpha}$.
Since the acid is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$,and $K_a \approx C \alpha^2$.
For condition $1$: $C_1 = 0.05 \ M$,$\alpha_1 = 0.10$.
$K_a = 0.05 \times (0.1)^2 = 0.05 \times 0.01 = 0.0005$.
For condition $2$: $C_2 = 0.10 \ M$,$\alpha_2 = ?$.
$K_a = C_2 \alpha_2^2 \implies 0.0005 = 0.10 \times \alpha_2^2$.
$\alpha_2^2 = \frac{0.0005}{0.10} = 0.005$.
$\alpha_2 = \sqrt{0.005} \approx 0.0707$.
Percentage dissociation $= 0.0707 \times 100 = 7.07 \%$.
The closest option is $7.17 \%$.

(B) The degree of dissociation $(\alpha)$ is calculated using the formula: $\alpha = \frac{\Lambda_m^c}{\Lambda_m^0}$
Given: $\Lambda_m^c = 16.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\Lambda_m^0 = 390.7 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Substituting the values: $\alpha = \frac{16.5}{390.7} = 0.04223$
Therefore,the degree of dissociation is approximately $0.0422$.

(C) The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a specific concentration $(\Lambda_c)$ to the molar conductivity at infinite dilution $(\Lambda_0)$:
$\alpha = \frac{\Lambda_c}{\Lambda_0}$
Given: $\Lambda_c = 3.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\Lambda_0 = 132 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\alpha = \frac{3.3}{132} = 0.025$
Percentage dissociation $= \alpha \times 100 = 0.025 \times 100 = 2.5 \%$.

(D) The molar conductivity of weak electrolytes increases as the concentration of the solution decreases.
This is because the degree of dissociation of the weak electrolyte increases with dilution.
Therefore,among the given concentrations,the lowest concentration,$0.001 \ M$,will exhibit the maximum molar conductivity.

(B) Given: $K_{a} = 1.8 \times 10^{-5}$ and $\alpha = 0.02$.
For a weak monobasic acid,the relationship between dissociation constant $(K_{a})$,degree of dissociation $(\alpha)$,and molar concentration $(c)$ is given by the formula:
$K_{a} = \alpha^2 c$
Rearranging the formula to solve for concentration $(c)$:
$c = \frac{K_{a}}{\alpha^2}$
Substituting the given values:
$c = \frac{1.8 \times 10^{-5}}{(0.02)^2} = \frac{1.8 \times 10^{-5}}{4 \times 10^{-4}} = 0.45 \times 10^{-1} \ M = 4.5 \times 10^{-2} \ M$.

(C) strong electrolyte is a substance that dissociates completely into ions in an aqueous solution.
$CH_3COOH$ (acetic acid) is a weak acid and a weak electrolyte.
$NH_4OH$ (ammonium hydroxide) is a weak base and a weak electrolyte.
$H_2CO_3$ (carbonic acid) is a weak acid and a weak electrolyte.
$NH_4Cl$ (ammonium chloride) is a salt formed from a strong acid $(HCl)$ and a weak base $(NH_4OH)$,and it dissociates completely into $NH_4^+$ and $Cl^-$ ions in water,making it a strong electrolyte.

(D) For a weak electrolyte $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant $K_{a}$ is given by $K_{a} = \frac{C\alpha^2}{1-\alpha}$.
Assuming $\alpha \ll 1$,we have $K_{a} \approx C\alpha^2$.
From this,$\alpha = \sqrt{\frac{K_{a}}{C}}$.
Since concentration $C = \frac{1}{V}$,where $V$ is the volume,we can substitute $C$ to get $K_{a} = \frac{\alpha^2}{V}$ or $\alpha = \sqrt{K_{a}V}$.
Comparing these with the given options:
Option $A$ is $\alpha = \sqrt{\frac{K_{a}}{C}}$ (Correct).
Option $B$ is $K = \frac{\alpha^2}{V}$ (Correct).
Option $C$ is $K = \alpha^2 C$ (Correct,as $K = C\alpha^2$).
Option $D$ is $\alpha = \sqrt{\frac{K_{a}}{V}}$,which is incorrect because $\alpha = \sqrt{K_{a}V}$.

(B) For a weak acid,the relationship between dissociation constant $(K_a)$,degree of dissociation $(\alpha)$,and concentration $(c)$ is given by $K_a = c \alpha^2$.
Given: $K_a = 1.8 \times 10^{-5}$ and $\alpha = 0.03$.
Rearranging the formula to solve for concentration: $c = \frac{K_a}{\alpha^2}$.
Substituting the values: $c = \frac{1.8 \times 10^{-5}}{(0.03)^2} = \frac{1.8 \times 10^{-5}}{9 \times 10^{-4}}$.
Calculating the result: $c = 0.2 \times 10^{-1} \ M = 0.02 \ M$.

(A) The concentration of $[H_3O^{+}]$ ions in a weak acid solution is given by the formula: $[H_3O^{+}] = \alpha \times c$
Given:
Concentration $(c) = 0.001 \ M = 10^{-3} \ M$
Degree of dissociation $(\alpha) = 0.134$
Substituting the values:
$[H_3O^{+}] = 0.134 \times 0.001$
$[H_3O^{+}] = 1.34 \times 10^{-4} \ mol \ L^{-1}$

(D) For a weak acid like $CH_3COOH$,the concentration $C = 0.1 \ N$ and dissociation constant $K_a = 1 \times 10^{-5}$.
Using the Ostwald dilution law,the degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.
Substituting the values: $\alpha = \sqrt{\frac{1 \times 10^{-5}}{0.1}} = \sqrt{1 \times 10^{-4}} = 1 \times 10^{-2}$.
Thus,the degree of dissociation is $1 \times 10^{-2}$.