Explanation

Correct answer is C

$\underset{\text{ (wA) }}{{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}}+\underset{\text{ (SB) }}{\mathrm{NaOH}}⟶\underset{\text{ (Salt) }}{{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}}+{\mathrm{H}}_2\mathrm{O}$


(A) $\to$ (B) Free ${\mathrm{H}}^{+}$ions are replaced by ${\mathrm{Na}}^{\oplus }$ which decreases conductance.

(B) $\to$ (C) Un-dissociated benzoic acid reacts with $\mathrm{NaOH}$ and forms salt which increases ions & conductance increases.

(C) $\to$ (D) After equivalence point at (3), $\mathrm{NaOH}$ added further increases ${\mathrm{Na}}^{\oplus }\mathrm{\&}{\mathrm{OH}}^{\odot }$ ions which further increases the conductance.

correct answer is 161

# Explanation

Using the Faraday's law of electrolysis, we can directly relate the amount of substance deposited (in this case, the nickel layer) with the electric charge passed through the electrolyte.

The formula for Faraday's law of electrolysis is:

$W=z\times i\times t$

where W is the amount of substance deposited (in grams), z is the electrochemical equivalent (grams per coulomb), i is the current (in amperes), and t is the time (in seconds).

By relating the density and volume of the nickel layer to the electric charge passed through the electrolyte, we can calculate the time needed for the deposition:

$10\times 100\times 0.0001=\frac{\left(\frac{\text{ atomic wt. }}{\text{ v.f }}\right)\times 2\times x}{96500}$

where v.f is the valence factor for the reaction (in this case, 2).

Solving for x, we get:

$x=161\sec$

So, the value of x is 161 seconds.

Explanation

Correct answer is D

In volumetric analysis, the concentration of a solution is a crucial factor in determining the accuracy of the results. In the case of an aqueous solution of KOH, the concentration can change over time due to the absorption of atmospheric CO₂. This occurs because KOH is a basic (alkaline) solution and reacts with carbon dioxide (CO₂) from the air to form potassium carbonate (K₂CO₃).

The reaction between KOH and CO₂ is given by :

KOH + CO₂$\to$ K₂CO₃

This reaction will change the concentration of KOH in the solution, which in turn will impact the accuracy of the results obtained in volumetric analysis. For example, if the concentration of KOH is lower than it was when the solution was prepared, then more solution will be needed to reach the endpoint in a reaction, and the results will be inaccurate.

That's why it is important to check the concentration of the KOH solution before using it for volumetric analysis, as stated in the assertion (A). And the reason (R) provides the explanation for why the concentration should be checked. Hence, both the assertion and the reason are correct and the correct answer is (D). Both (A) and (R) are correct and (R) is the correct explanation of (A).

Explanation

Correct answer is D

${\mathrm{H}}_2{\mathrm{O}}_{(l)}+e^{-}⟶\frac{1}{2}{\mathrm{H}}_{2(g)}+{\mathrm{OH}}_{(aq)}^{-}$

Anode $:{\mathrm{Cl}}_{(aq)}^{-}⟶\frac{1}{2}{\mathrm{Cl}}_{2(g)}+e^{-}$

correct answer is 3

# Explanation

For ${\mathrm{Fe}}_3{\mathrm{O}}_4$,

$x=\frac{+8}{3}$

where x is oxidation state of Fe.

${\mathrm{Fe}}_3{\mathrm{O}}_4+8{\mathrm{H}}^{+}+8{\mathrm{e}}^{-}⟶3\mathrm{Fe}+4{\mathrm{H}}_2\mathrm{O}$

Charge required $=\frac{8}{3}\times \mathrm{F}=\frac{8\mathrm{F}}{3}\simeq 3\mathrm{F}$

correct answer is 127

# Explanation

$2\mathrm{F}$ produces $=\frac{3}{2}$ mole of gas

$0.10\times 2\times 3600$ coulomb produces

$\begin{array}{rl}& =\frac{\frac{3}{2}\times 0.1\times 2\times 3600}{2\times 96500}\\ \\ & =0.0056\text{ moles of gas }\end{array}$

Volume of gas produced $=0.0056\times 22.7\mathrm{L}$

$\begin{array}{rl}& \simeq 0.127\mathrm{L}\\ \\ & =127\mathrm{mL}\end{array}$

correct answer is 6

# Explanation

$\stackrel{+6}{{\mathrm{Cr}}_2}{\mathrm{O}}_7^{2-}⟶2{\mathrm{Cr}}^{3+}$

$\because$ Each $\mathrm{Cr}$ is converting from $+6$ to $+3$

$\therefore 6$ faradays of charge is required

correct answer is 20

# Explanation

${\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe}$

$\text{Moles of Fe deposited }=\frac{0.3482}{56}=6.2\times {10}^{-3}$

For 1 mole $\mathrm{Fe}$, charge required is $3\mathrm{F}$

For $6.2\times {10}^{-3}$ mole $\mathrm{Fe}$, charge required is $3\times 6.2\times {10}^{-3}\mathrm{F}$

Since, charge required $=18.6\times {10}^{-3}\times 96500\mathrm{C}$

$=1794.9\mathrm{C}$

And,

$1.5\times \mathrm{t}=1794.9$

$t=\frac{1794.9}{1.5\times 60}min$

$\mathrm{t}\simeq 20\min$

correct answer is 1

# Explanation

$W=\frac{E}{F}\times I\times t$

$10=\frac{122.6}{96500\times 6}\times x\times 10\times 3600$

$x=1.311$

Ans. (1)