A) \[4.68\times {{10}^{18}}\]
B) \[4.68\times {{10}^{15}}\]
C) \[4.68\times {{10}^{12}}\]
D) \[4.68\times {{10}^{9}}\]
Correct Answer: A
Solution :
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, \[i=25\,mA=0.0025A\] \[t=60s\] \[Q=it\] \[=0\text{ }.0025\times 60\] \[=1.5C\] Number of electrons in 1.5 C \[=\frac{Q\times Avogadro\text{ }number}{96500}\] \[=\frac{1.5\times 6.023\times {{10}^{23}}}{96500}\] \[=9.36\times {{10}^{18}}\] \[Ca\xrightarrow[{}]{{}}C{{a}^{2+}}+2{{e}^{-}}\] \[\because \] \[2{{e}^{-}}\]are required to deposit 1 Ca atom. \[\therefore \]no. of Ca atoms deposited \[=\frac{no.\text{ }of\text{ }electrons}{2}\] \[=\frac{9.36\times {{10}^{18}}}{2}\] \[=4.68\times {{10}^{18}}\]
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