How many atoms of calcium will be deposited from a solution of $C a C l_{2}$ by a current of 25 miliampere flowing for 60 seconds?

4.68 \times 10^{18}

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4.68 \times 10^{15}

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4.68 \times 10^{12}

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4.68 \times 10^{9}

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Solution

The correct option is A $4.68 \times 10^{18}$
We know that :

$M = Z \times I \times t$

But, $Z = \frac{E q u i v a l e n t w e i g h t}{96500}$

Put the value of Z in above equation, we get :

$M = \frac{20}{2 \times 96500} \times 25 \times 10^{- 3} \times 60$

$M = 0.00015544$

No. of moles $= \frac{0.00015544}{20} = 0.000007772$

No. of atoms $= 0.000046811 \times 10^{23} = 4.6811 \times 10^{18}$

Hence the correct option is A.

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How many atoms of calcium will be deposited from a solution of CaCl2 by a current of 25 miliampere flowing for 60 seconds?

The correct option is A 4.68×1018We know that :M=Z×I×t But, Z=Equivalentweight96500Put the value of Z in above equation, we get :M=202×96500×25×10−3×60M=0.00015544No. of moles =0.0001554420=0.000007772No. of atoms =0.000046811×1023=4.6811×1018Hence the correct option is A.

How many atoms of calcium will be deposited from a solution of $C a C l_{2}$ by a current of 25 miliampere flowing for 60 seconds?