Electrical conductors and Ostwald’s dilution law Questions in English

(D) The correct answer is $D$.
Electrical conductivity in aqueous solutions is due to the presence of free ions.
$NaCl$,$CuSO_4$,and $NH_4Cl$ are electrolytes that dissociate into ions in water,making the solution conducting.
Sugar $(C_{12}H_{22}O_{11})$ is a non-electrolyte and does not dissociate into ions in water; therefore,it does not make water conducting.

(D) Urea $(NH_2CONH_2)$ is a covalent compound.
When dissolved in water,it does not dissociate into ions.
Since electricity conduction in aqueous solutions requires the presence of free ions,an aqueous solution of urea does not conduct electricity.

(B) Ammonium hydroxide $(NH_4OH)$ is a weak base.
It dissociates only partially in aqueous solution into $NH_4^+$ and $OH^-$ ions.
Therefore,it is classified as a weak electrolyte.

(B) According to Ostwald's dilution law,the degree of dissociation $(\alpha)$ of a weak electrolyte is related to the concentration $(C)$ by the expression $\alpha = \sqrt{\frac{K_a}{C}}$.
Thus,the degree of dissociation is inversely proportional to the square root of the concentration of the electrolyte.
Therefore,decreasing the concentration of the electrolyte increases the degree of dissociation.

(D) strong electrolyte is a substance that dissociates completely into ions in an aqueous solution.
$(A)$ Urea $(NH_2CONH_2)$ is a non-electrolyte.
$(B)$ Ammonium hydroxide $(NH_4OH)$ is a weak electrolyte.
$(C)$ Sugar $(C_{12}H_{22}O_{11})$ is a non-electrolyte.
$(D)$ Sodium acetate $(CH_3COONa)$ is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$. Salts are generally strong electrolytes because they dissociate completely in water: $CH_3COONa(aq) \rightarrow CH_3COO^{-}(aq) + Na^{+}(aq)$.

(A) $NaCl$ is a strong electrolyte because it is an ionic salt that dissociates completely into its constituent ions in an aqueous solution.
$CH_3COOH$ and $NH_4OH$ are weak electrolytes as they undergo partial dissociation.
$C_6H_{12}O_6$ (glucose) is a non-electrolyte as it does not dissociate into ions.

(C) For a weak acid $HA$ dissociating as $HA \rightleftharpoons H^+ + A^-$,the equilibrium constant $K_a$ is given by $K_a = \frac{[H^+][A^-]}{[HA]}$.
At equilibrium,$[H^+] = C\alpha$,$[A^-] = C\alpha$,and $[HA] = C(1 - \alpha)$.
Substituting these,$K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$.
Since the electrolyte is weak,$\alpha \ll 1$,so $(1 - \alpha) \approx 1$.
Thus,$K_a \approx C\alpha^2$,which simplifies to $\alpha^2 = \frac{K_a}{C}$ or $\alpha = \sqrt{\frac{K_a}{C}}$.

(B) According to Ostwald's dilution law,the degree of ionization $(\alpha)$ is inversely proportional to the square root of the concentration $(C)$ of the electrolyte,i.e.,$\alpha = \sqrt{K_a/C}$.
As the amount of water added increases,the concentration $(C)$ decreases,which leads to an increase in the degree of ionization $(\alpha)$.
Therefore,the extent of ionization increases on the addition of excess water to the solution.

(D) At infinite dilution,the concentration of the electrolyte approaches zero.
According to Ostwald's dilution law,the degree of ionization $(\alpha)$ increases with dilution.
As the dilution approaches infinity,the degree of ionization for any electrolyte (whether strong or weak) approaches $1$,which corresponds to $100 \%$ ionization.

(B) The degree of ionization of a solute depends upon its nature,concentration,and temperature.

(B) For a weak acid $HA$ dissociating as $HA \rightleftharpoons H^+ + A^-,$ the initial concentration is $c$ and the degree of dissociation is $\alpha.$
At equilibrium,the concentrations are $[HA] = c(1 - \alpha),$ $[H^+] = c\alpha,$ and $[A^-] = c\alpha.$
The dissociation constant $K_a$ is given by $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(c\alpha)(c\alpha)}{c(1 - \alpha)} = \frac{c^2\alpha^2}{c(1 - \alpha)} = \frac{c\alpha^2}{1 - \alpha}.$
Thus,the correct expression is $K_a = \frac{\alpha^2 c}{1 - \alpha}.$

(C) According to Ostwald's dilution law,the degree of ionization $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.
Since dilution $V = \frac{1}{C}$,we can substitute $C$ with $\frac{1}{V}$ to get $\alpha = \sqrt{K_a \cdot V}$.
Therefore,for a weak electrolyte,the degree of ionization $\alpha$ is proportional to the square root of the dilution $V$ (i.e.,$\alpha \propto \sqrt{V}$).

(C) According to Ostwald's dilution law,the degree of ionisation $(\alpha)$ of a weak electrolyte is related to the dilution $(V)$ by the expression $\alpha = \sqrt{K_a \times V}$.
Thus,ionisation increases as dilution increases.

(A) According to Ostwald's dilution law,for a weak electrolyte,the degree of dissociation $(\alpha)$ is related to the dilution $(V)$ by the expression $\alpha = \sqrt{K_a \times V}$.
As the dilution $(V)$ increases,the degree of dissociation $(\alpha)$ increases.

(C) For a weak acid $HA \rightleftharpoons H^+ + A^-$,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{K_a / C}$.
As the concentration $C$ decreases (dilution),the degree of dissociation $\alpha$ increases.
Since $[H^+] = C \times \alpha = \sqrt{K_a \times C}$,as $C$ decreases,$[H^+]$ also decreases.
However,the decrease in $[H^+]$ is not proportional to the decrease in $C$ because $\alpha$ increases.
Since $[H^+]$ decreases,the $pH = -\log[H^+]$ will increase.
Therefore,the percentage ionization increases upon dilution,while the $pH$ increases (not decreases).

(A) For a weak acid $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant $K_a$ is given by $K_a = \frac{[H^{+}][A^{-}]}{[HA]}$.
Upon dilution,the degree of dissociation $(\alpha)$ increases according to Ostwald's dilution law,$\alpha = \sqrt{\frac{K_a}{C}}$.
As $\alpha$ increases,$[H^{+}]$ changes,and consequently,the $pH$ changes.
However,the equilibrium constant $K_a$ is a characteristic property of the acid at a constant temperature and does not change upon dilution.

(B) . In an aqueous solution,$HCl$ undergoes ionisation to produce $H^{+}_{(aq)}$ and $Cl^{-}_{(aq)}$ ions,which act as charge carriers. The reaction is: $HCl \xrightarrow{H_2O} H^{+}_{(aq)} + Cl^{-}_{(aq)}$. In the gaseous state,$HCl$ remains as covalent molecules and does not provide ions for conduction.

(B) . Sugar $(C_{12}H_{22}O_{11})$ is a covalent compound that does not dissociate into ions in water. Since electrical conductivity in aqueous solutions requires the presence of mobile ions,sugar solution does not conduct electricity.

(D) Strong electrolytes are substances that undergo complete dissociation into their constituent ions in an aqueous solution,regardless of the concentration or dilution level.
Therefore,they exist almost entirely as ions in the solution.

(C) Strong electrolytes are substances that dissociate or ionise almost completely into ions when dissolved in a polar solvent like water. Therefore,option $(C)$ is correct.

(B) $HCl$ is a strong electrolyte that dissociates completely into $H^+$ and $Cl^-$ ions in an aqueous solution.
These free ions are responsible for the conduction of electric current.
Glycerol,sugar,and pure water are non-electrolytes or very poor conductors of electricity.

(D) The electric conductivity of a solution is directly proportional to the number of charge carriers (ions) present in it.
As the extent of ionization increases,the number of ions in the solution increases.
Therefore,the electric conduction of a salt solution depends on the extent of its ionization.

(B) The electrical conductivity of a substance depends on the presence of mobile ions or free electrons.
Pure water $(H_2O)$ is a very weak electrolyte and is almost totally unionized into $H^+$ and $OH^-$ ions.
Due to the extremely low concentration of these ions,it does not conduct electricity significantly.
Therefore,the correct option is $(b)$.

(B) Gaseous $HCl$ is a covalent compound and does not contain free ions.
When $HCl$ is dissolved in water,it undergoes ionization to produce $H^{+}_{(aq)}$ and $Cl^{-}_{(aq)}$ ions.
The presence of these mobile ions allows the solution to conduct electricity.
The reaction is: $HCl_{(g)} \stackrel{H_2O}{\longrightarrow} H^{+}_{(aq)} + Cl^{-}_{(aq)}$.

(B) $C_2H_5OH$ (Ethanol) is a covalent compound and a non-electrolyte.
It does not dissociate into ions in an aqueous solution,which is necessary for the conduction of electricity.
Therefore,it is a poor conductor of electricity.

(C) According to $Ostwald's \ dilution \ law$,the degree of ionization $(\alpha)$ of a weak electrolyte is given by $\alpha = \sqrt{K_a / C}$,where $K_a$ is the dissociation constant and $C$ is the concentration of the electrolyte.
Since $\alpha \propto 1 / \sqrt{C}$,as the concentration $C$ decreases (which happens upon adding more water or dilution),the degree of ionization $\alpha$ increases.

(A) Ostwald's dilution law is derived based on the assumption of an equilibrium between the undissociated molecules and the ions of a weak electrolyte,represented as $AB \rightleftharpoons A^+ + B^-$.
For strong electrolytes,the dissociation is essentially complete at all concentrations,meaning the degree of dissociation $\alpha \approx 1$.
Since there is no significant equilibrium between the undissociated solute and its ions,the law of mass action cannot be applied in the same way as it is for weak electrolytes.

(B) According to Ostwald's dilution law,the degree of dissociation $\alpha$ for a weak electrolyte is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$,where $K_a$ is the dissociation constant and $C$ is the molar concentration.
Since $C = \frac{n}{V}$,where $V$ is the volume of the solution,we have $\alpha = \sqrt{K_a \times V}$.
As the dilution increases,the volume $V$ increases,which leads to an increase in the degree of dissociation $\alpha$.

(A) The dissociation constant for a weak electrolyte is given by the formula: $K = C\alpha^2$.
Rearranging for the degree of dissociation $\alpha$: $\alpha = \sqrt{\frac{K}{C}}$.
Substituting the given values: $\alpha = \sqrt{\frac{4.41 \times 10^{-5}}{0.1}}$.
$\alpha = \sqrt{4.41 \times 10^{-4}}$.
$\alpha = 2.1 \times 10^{-2}$.

(B) strong electrolyte is a substance that completely dissociates into its constituent ions in an aqueous solution.
$NH_4OH$ is a weak base.
$HCN$ is a weak acid.
$H_2SO_3$ is a weak acid.
$Ca(NO_3)_2$ is a salt of a strong acid $(HNO_3)$ and a strong base $(Ca(OH)_2)$,which dissociates completely in water as $Ca(NO_3)_2(aq) \rightarrow Ca^{2+}(aq) + 2NO_3^-(aq)$.
Therefore,$Ca(NO_3)_2$ is a strong electrolyte.

(B) The degree of ionization $(\alpha)$ of a compound is primarily determined by the nature of the solute (the electrolyte) and the nature of the solvent. It also depends on factors like temperature and concentration. Among the given options,the nature of the solute molecule is the most significant factor affecting the extent of dissociation.

(A) $Ostwald's$ dilution law describes the relationship between the dissociation constant of a weak electrolyte and its degree of dissociation. Therefore,the name $Ostwald$ is associated with electrolytic dissociation.

(B) The degree of dissociation $(\alpha)$ is given by the ratio of equivalent conductance at a specific concentration $(\Lambda_{eq})$ to the equivalent conductance at infinite dilution $(\Lambda^0_{eq})$.
Given that $\Lambda_{eq} = \frac{\Lambda^0_{eq}}{100}$.
Therefore,$\alpha = \frac{\Lambda_{eq}}{\Lambda^0_{eq}} = \frac{1}{100} = 0.01$.

(C) strong electrolyte is defined as a substance that dissociates completely into its constituent ions in an aqueous solution,even at high concentrations.

(A) $HCl$ is a strong acid and dissociates completely into ions in aqueous solution.
Since electrical conductivity depends on the number of free ions present in the solution,$HCl$ acts as the best conductor among the given options.
$NH_3$ and $CH_3COOH$ are weak electrolytes,while $C_6H_{12}O_6$ is a non-electrolyte.

(D) Degree of dissociation,$\alpha = \frac{\Lambda^{c}}{\Lambda^{\infty}}$
Given,$\Lambda^{c} = 8.0 \, \text{mho} \, \text{cm}^2$ and $\Lambda^{\infty} = 400 \, \text{mho} \, \text{cm}^2$.
$\alpha = \frac{8.0}{400} = 0.02 = 2 \times 10^{-2}$.
For a weak monobasic acid,the dissociation constant $K_{a}$ is given by Ostwald's dilution law:
$K_{a} = \frac{C \alpha^{2}}{1 - \alpha}$.
Since $\alpha$ is very small $(1 - \alpha \approx 1)$,the formula simplifies to $K_{a} \approx C \alpha^{2}$.
Given concentration $C = \frac{1}{32} \, M$.
$K_{a} = \frac{1}{32} \times (2 \times 10^{-2})^{2} = \frac{1}{32} \times 4 \times 10^{-4} = \frac{1}{8} \times 10^{-4} = 0.125 \times 10^{-4} = 1.25 \times 10^{-5}$.

(A) According to Ostwald's dilution law,the dissociation constant $K$ for a weak electrolyte is given by:
$K = \frac{C \alpha^2}{1 - \alpha}$
Rearranging the equation:
$K(1 - \alpha) = C \alpha^2$
$K - K \alpha = C \alpha^2$
Bringing all terms to one side to form a quadratic equation:
$C \alpha^2 + K \alpha - K = 0$

(D) For a weak electrolyte,the degree of dissociation $(\alpha)$ increases with dilution according to Ostwald's dilution law.
As dilution increases,the number of ions per unit volume increases due to the increase in the degree of dissociation,which leads to an increase in molar conductivity $(\Lambda_m)$.
At infinite dilution,the electrolyte dissociates completely,but the concentration of ions becomes extremely low,making the conductivity $(\kappa)$ very small and difficult to measure accurately.
Therefore,option $D$ is the correct statement.

(D) The degree of ionisation $(\alpha)$ is calculated using the formula: $\alpha = \frac{\lambda^{c}}{\lambda^{\infty}}$
Given,$\lambda^{c} = 9.54\, \Omega^{-1}\, cm^2\, mol^{-1}$ and $\lambda^{\infty} = 238\, \Omega^{-1}\, cm^2\, mol^{-1}$.
$\alpha = \frac{9.54}{238} = 0.04008$
To express this as a percentage: $\% \alpha = 0.04008 \times 100 = 4.008\, \%$

(B) For a weak acid,the degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.
Given $C_1 = 0.1 \ M$ and $C_2 = 0.001 \ M$.
The ratio of the degrees of dissociation is $\frac{\alpha_1}{\alpha_2} = \sqrt{\frac{C_2}{C_1}}$.
Substituting the values: $\frac{\alpha_1}{\alpha_2} = \sqrt{\frac{10^{-3}}{10^{-1}}} = \sqrt{10^{-2}} = 10^{-1} = 0.1$.

(C) substance conducts electricity in an aqueous solution only if it dissociates into ions.
Ionic compounds like $FeSO_4$ (Green vitriol),$KNO_3$ or $NaNO_3$ (Saltpetre),and $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ (Potash alum) dissociate into ions in water and are therefore conducting.
Alcohol (e.g.,$C_2H_5OH$) is a covalent compound that does not dissociate into ions when dissolved in water.
Therefore,an aqueous solution of alcohol is non-conducting.

(B) For a weak monobasic acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.
In the first case,$C_1 = 0.1 \ M$ and $\alpha_1 = 0.01$ (since $1\%$ ionised).
So,$K_a = C_1 \alpha_1^2 = 0.1 \times (0.01)^2 = 10^{-5}$.
In the second case,$C_2 = 0.025 \ M$.
The new degree of dissociation $\alpha_2 = \sqrt{\frac{K_a}{C_2}} = \sqrt{\frac{10^{-5}}{0.025}} = \sqrt{\frac{10^{-5}}{2.5 \times 10^{-2}}} = \sqrt{4 \times 10^{-4}} = 0.02$.
Therefore,the percentage ionisation is $0.02 \times 100 = 2\%$.

(B) The equivalent conductance at concentration $c$ is given as $\lambda_{eq}^{c} = 12.8 \ ohm^{-1} \ cm^{2} \ eq^{-1}$.
According to Kohlrausch's law,the equivalent conductance at infinite dilution is the sum of the ionic conductances of the constituent ions:
$\lambda_{eq}^{\infty}(C_{6}H_{5}COOH) = \lambda_{eq}^{\infty}(C_{6}H_{5}COO^{-}) + \lambda_{eq}^{\infty}(H^{+})$
Substituting the given values:
$\lambda_{eq}^{\infty} = 42 + 288.42 = 330.42 \ ohm^{-1} \ cm^{2} \ eq^{-1}$.
The degree of dissociation $\alpha$ is calculated as:
$\alpha = \frac{\lambda_{eq}^{c}}{\lambda_{eq}^{\infty}} = \frac{12.8}{330.42} \approx 0.03874$.
To express this as a percentage:
$\% \alpha = 0.03874 \times 100 \approx 3.874 \% \approx 3.9 \%$.

(C) According to Ostwald's dilution law,the degree of dissociation $(\alpha)$ of a weak electrolyte is related to the dilution $(V)$ by the expression $\alpha = \sqrt{K_a \cdot V}$,where $K_a$ is the dissociation constant.
As the dilution $(V)$ increases (which means the concentration decreases),the degree of dissociation $(\alpha)$ increases.
Therefore,increasing the dilution leads to an increase in the degree of dissociation.

(B) For a weak acid $HA$,the dissociation equilibrium is: $HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
At equilibrium,the concentrations are: $C(1 - \alpha)$,$C\alpha$,and $C\alpha$ respectively.
The expression for the ionisation constant is $K_a = \frac{C\alpha^2}{1 - \alpha}$.
The degree of dissociation $\alpha$ is given by the ratio of molar conductivity at concentration $C$ to molar conductivity at infinite dilution: $\alpha = \frac{\Lambda _m}{\Lambda _m^\infty}$.
Substituting $\alpha$ into the $K_a$ expression:
$K_a = \frac{C(\frac{\Lambda _m}{\Lambda _m^\infty})^2}{1 - \frac{\Lambda _m}{\Lambda _m^\infty}}$
Simplifying the expression:
$K_a = \frac{C\Lambda _m^2}{(\Lambda _m^\infty)^2} \times \frac{\Lambda _m^\infty}{(\Lambda _m^\infty - \Lambda _m)}$
$K_a = \frac{C\Lambda _m^2}{\Lambda _m^\infty (\Lambda _m^\infty - \Lambda _m)}$

(C) For a weak base,the degree of dissociation $(\alpha)$ is given by the formula $\alpha = \sqrt{\frac{K_b}{C}}$,where $K_b$ is the dissociation constant of the base and $C$ is the molar concentration.
From the relation $\alpha \propto \frac{1}{\sqrt{C}}$,it is clear that the degree of dissociation increases as the concentration $(C)$ decreases.
Comparing the given concentrations: $1 \ M$,$0.1 \ M$,$0.01 \ M$,and $0.02 \ M$.
The lowest concentration is $0.01 \ M$.
Therefore,the degree of dissociation will be the highest for the $0.01 \ M$ solution.

(C) Ostwald's dilution law is applicable only to weak electrolytes.
$HCl$,$HNO_3$,and $NaOH$ are strong electrolytes that dissociate completely in solution.
$CH_3COOH$ (acetic acid) is a weak electrolyte that dissociates partially in solution.
Therefore,Ostwald's dilution law gives satisfactory results with $CH_3COOH$.

(D) At infinite dilution,the concentration of the electrolyte approaches zero.
According to Ostwald's dilution law and the concept of molar conductivity,as the dilution increases,the degree of dissociation $(\alpha)$ increases.
At infinite dilution,the electrolyte dissociates completely into its constituent ions.
Therefore,the percentage ionization becomes $100\%$ for all electrolytes.