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Chemistry · 2023 · 5 marks
CBSE 2023 · Region 2 · Set 1 · Q35
(a) Conductivity of 2 × 10^-3 M methanoic acid is 8 × 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation if Λ_m^o for…
(a)
Conductivity of $\displaystyle 2 \times 10^{-3} \mathrm{M}$ methanoic acid is $\displaystyle 8 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and degree of dissociation if $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\mathrm{o}}$ for methanoic acid is $\displaystyle 404 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. $\displaystyle 3$ + $\displaystyle 2$
(b)
Calculate the $\displaystyle \Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ and $\displaystyle \log \mathrm{K}_{\mathrm{c}}$ for the given reaction at $\displaystyle 298$ K: \[\mathrm{Ni}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})}^{+} \rightleftharpoons \mathrm{Ni}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{Ag}_{(\mathrm{s})} \] Given: $\displaystyle \mathrm{E}^{\circ}{ }_{\mathrm{N}_{\mathrm{i}}}{ }^{2+} / \mathrm{N}_{\mathrm{i}}=-0.25 \mathrm{~V}, \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=+0.80 \mathrm{~V}$ \[1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1} . \]
Marking-scheme solution
(a)
\[\begin{aligned}
\mathbf{\Lambda}_{\mathrm{m}} & =\frac{k}{\mathrm{c}} \times 1000 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\
& =\frac{8 \times 10^{-5}}{2 \times 10^{-3}} \times 1000 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\
& =40 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\
\alpha & =\frac{\mathbf{\Lambda}_{\mathrm{m}}}{\mathbf{\Lambda}_{\mathrm{m}}{ }^{\circ}} \\
& =\frac{40}{404} \\
& =0.099
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.