How many atoms of calcium will be deposited f | vedclass

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(B) Given,Current $(i) = 0.25 \, mA = 0.25 	imes 10^{-3} \, A$. Time $(t) = 60 \, s$. Charge $(Q) = i 	imes t = 0.25 	imes 10^{-3} 	imes 60 = 0.01

Vedclass · Accepted Answer

$4.68 	imes 10^{16}$

Vedclass · Answer

(B) Given,Current $(i) = 0.25 \, mA = 0.25 	imes 10^{-3} \, A$. Time $(t) = 60 \, s$. Charge $(Q) = i 	imes t = 0.25 	imes 10^{-3} 	imes 60 = 0.015 \, C$. Number of electrons $(n_e) = \frac{Q}{F} = \frac{0.015}{96500} \approx 1.554 	imes 10^{-7} \, 	ext{moles of electrons}$. Number of electrons $= 1.554 	imes 10^{-7} 	imes 6.022 	imes 10^{23} \approx 9.36 	imes 10^{16} \, 	ext{electrons}$. Reaction: $Ca^{2+} + 2e^- 	o Ca$. Since $2$ moles of electrons deposit $1$ mole of $Ca$ atoms,the number of $Ca$ atoms $= \frac{9.36 	imes 10^{16}}{2} = 4.68 	imes 10^{16}$.

How many atoms of calcium will be deposited from a solution of $CaCl_2$ by a current of $0.25 \, mA$ flowing for $60 \, s$?

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