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Choose the Best Answer · Q1

Q.The number of electrons that have a total charge of 9650 coulombs is

(a) 6.22×10236.22\times10^{23}
(b) 6.022×10246.022\times10^{24}
(c) 6.022×10226.022\times10^{22}
(d) 6.022×10−346.022\times10^{-34}
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✓ Free question

Step 1. The charge carried by a single electron is e=1.602×10−19e = 1.602\times10^{-19} C, so the number of electrons carrying a total charge Q is simply n=Q/en = Q/e.

Step 2. Substituting Q=9650Q = 9650 C: n=96501.602×10−19=6.024×1022n = \dfrac{9650}{1.602\times10^{-19}} = 6.024\times10^{22} electrons.

Step 3. Comparing with the options, this matches (c). Note that one mole of electrons (6.022×10236.022\times10^{23}) would carry a full faraday, 96500 C — since 9650 C is exactly one-tenth of a faraday, the electron count is correspondingly one-tenth of Avogadro's number, confirming the order of magnitude.

✓Final answer

(c) 6.022×10226.022\times10^{22} electrons.

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