Question

How many atoms of calcium will be deposited from a solution of $ CaC{{l}_{2}} $ by a current of 5 mA flowing for 60s?

• A. $ 4.68\times {{10}^{18}} $
• B. $ 4.68\times {{10}^{15}} $
• C. $ 4.68\times {{10}^{12}} $
• D. $ 4.68\times {{10}^{9}} $

Answer 1

Answer: (A)

Key Idea: First calculate total current passed (Q) by equating it to as follows Q = it then number of electrons and then number of atoms of Ca deposited. Given, $ i=25mA=0.0025A $ $ t=60\text{ }s $ Q = it $ =0.0025\times 60=1.5C $ Number of electrons in 1.5 C $ =\frac{Q\times Avogadro\,number}{96500} $ $ =\frac{1.5\times 6.023\times {{10}^{23}}}{96500} $ $ =9.36\times {{10}^{18}} $ $ Ca\xrightarrow{{}}C{{a}^{2+}}+2{{e}^{-}} $ $ \because $ $ 2{{e}^{-}} $ are required to deposit 1 Ca atom. $ \therefore $ No. of Ca atoms deposited $ =\frac{no.\,\,of\,electrons}{2} $ $ =\frac{9.36\times {{10}^{18}}}{2} $ $ =4.68\times {{10}^{18}} $