Identify the correct statements from the following:
$(A)$ At $298 \ K$,the potential of a hydrogen electrode placed in a solution of $pH = 10$ is $-0.59 \ V$.
$(B)$ The limiting molar conductivity of $Ca^{2+}$ and $Cl^{-}$ are $119$ and $76 \ S \ cm^2 \ mol^{-1}$ respectively. The limiting molar conductivity of $CaCl_2$ is $195 \ S \ cm^2 \ mol^{-1}$.
$(C)$ The correct relationship between $K_{c}$ and $E_{cell}^{0}$ is $E_{cell}^{0} = \frac{2.303 RT}{nF} \log K_{c}$.
$(A)$ At $298 \ K$,the potential of a hydrogen electrode placed in a solution of $pH = 10$ is $-0.59 \ V$.
$(B)$ The limiting molar conductivity of $Ca^{2+}$ and $Cl^{-}$ are $119$ and $76 \ S \ cm^2 \ mol^{-1}$ respectively. The limiting molar conductivity of $CaCl_2$ is $195 \ S \ cm^2 \ mol^{-1}$.
$(C)$ The correct relationship between $K_{c}$ and $E_{cell}^{0}$ is $E_{cell}^{0} = \frac{2.303 RT}{nF} \log K_{c}$.
- A$A, B, C$
- B$A, B$ only
- C$A, C$ only
- D$B, C$ only
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An aqueous solution containing $6.5 \ g$ of $NaCl$ of $90 \%$ purity was subjected to electrolysis. After the complete electrolysis,the solution was evaporated to get solid $NaOH$. The volume of $1 \ M$ acetic acid required to neutralize $NaOH$ obtained above is (in $cm^{3}$)
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View SolutionThe standard reduction potentials of $2H^{+}/H_2$,$Cu^{2+}/Cu$,$Zn^{2+}/Zn$ and $NO_3^{-}, H^{+}/NO$ are $0.0$,$0.34$,$-0.76$ and $0.97 \ V$ respectively. Identify the correct statements from the following:
$I.$ $H^{+}$ does not oxidize $Cu$ to $Cu^{2+}$
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View SolutionMatch the column $I$ with column $II$ and mark the appropriate choice.
Column $I$ Column $II$ $A$. Kohlrausch law $i$. $\Lambda _{m}^o = \nu _+ \lambda _+^o + \nu _- \lambda _-^o$ $B$. Molar Conductivity $ii$. $\Lambda _m = \frac{\kappa \times 1000}{M}$ $C$. Degree of Dissociation $iii$. $\alpha = \frac{\Lambda _m}{\Lambda _m^o}$ $D$. Dissociation Constant $iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$
| Column $I$ | Column $II$ |
|---|---|
| $A$. Kohlrausch law | $i$. $\Lambda _{m}^o = \nu _+ \lambda _+^o + \nu _- \lambda _-^o$ |
| $B$. Molar Conductivity | $ii$. $\Lambda _m = \frac{\kappa \times 1000}{M}$ |
| $C$. Degree of Dissociation | $iii$. $\alpha = \frac{\Lambda _m}{\Lambda _m^o}$ |
| $D$. Dissociation Constant | $iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$ |
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View SolutionConsider the following data.
Electrolyte $\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$ $BaCl_2$ $x_1$ $H_2SO_4$ $x_2$ $HCl$ $x_3$
$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x\text{ S cm}^{-1}$, then the solubility product of $BaSO_4$ can be given as (Here $\Lambda_m = \Lambda_m^\circ$)
| Electrolyte | $\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$ |
|---|---|
| $BaCl_2$ | $x_1$ |
| $H_2SO_4$ | $x_2$ |
| $HCl$ | $x_3$ |
$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x\text{ S cm}^{-1}$, then the solubility product of $BaSO_4$ can be given as (Here $\Lambda_m = \Lambda_m^\circ$)
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