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6. ⇒ (MHT CET 2021 21th September Morning Shift )

What is the number of electrons passed through an electrolyte solution when 1 ampere current is passed for 16.1 minutes?

A. $5.022 \times 10^{24}$

B. $3.011 \times 10^{22}$

C. $6.022 \times 10^{21}$

D. $2.022 \times 10^{23}$

Correct Option is (C)

$\begin{aligned} I = 1 A, t = 16.1 \min = 16.1 \times 60 s \\ Moles of electrons actually passed & = \frac{Q (C)}{96500 (C / mol e^{- .})} \\ = \frac{I \times t}{96500} \\ = \frac{1 \times 16.1 \times 60}{96500} \\ = 0.01 mol \end{aligned}$

Now, $1 mol = 6.022 \times 10^{23}$ electrons

$∴ 0.01 mol = 6.022 \times 10^{21}$ electrons

7. ⇒ (MHT CET 2021 20th September Evening Shift )

How many faraday of electricity is required to produce 10 g of calcium metal (molar mass = 40 g mol $^{- 1}$ ) from calcium ions?

A. 1.5 F

B. 2.0 F

C. 0.50 F

D. 1.0 F

Correct Option is (C)

${Ca}_{(aq)}^{2 +} + 2 e^{-} ⟶ {Ca}_{(s)}$

$40 g$ of $Ca$ requires $2 F$ of electricity

$∴ 10 g of Ca = \frac{2 \times 10}{40} = 0.5 F of electricity$

⇒Electrochemistry