MHT CET · Chemistry · Electrochemistry
How many electrons are involved in the reaction when \(0 \cdot 40\) F of electricity is passed through an electrolytic solution?
- A \(6.642 \times 10^{25}\)
- B \(2.4088 \times 10^{23}\)
- C \(1.505 \times 10^{24}\)
- D \(6.022 \times 10^{23}\)
Answer & Solution
Correct Answer
(B) \(2.4088 \times 10^{23}\)
Step-by-step Solution
Detailed explanation
(C)
\(\begin{array}{l}
1 \mathrm{~F}=96500 \mathrm{C} \\
\therefore \quad 0.4 \mathrm{~F}=96500 \times 0.4=38600 \mathrm{C}
\end{array}\)
\(\begin{aligned}
\text { Now, } 96500 \mathrm{C} &=6.022 \times 10^{23} \text { electrons } \\
\therefore 38600 \mathrm{C} &=\frac{6.022 \times 10^{23} \times 38600}{96500} \\
&=2.4088 \times 10^{23} \text { electrons }
\end{aligned}\)
\(\begin{array}{l}
1 \mathrm{~F}=96500 \mathrm{C} \\
\therefore \quad 0.4 \mathrm{~F}=96500 \times 0.4=38600 \mathrm{C}
\end{array}\)
\(\begin{aligned}
\text { Now, } 96500 \mathrm{C} &=6.022 \times 10^{23} \text { electrons } \\
\therefore 38600 \mathrm{C} &=\frac{6.022 \times 10^{23} \times 38600}{96500} \\
&=2.4088 \times 10^{23} \text { electrons }
\end{aligned}\)
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