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, \[i=25mA=0.0025A\] \[t=60\text{ }s\] 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 \[=\frac{no.\,\,of\,electrons}{2}\] \[=\frac{9.36\times {{10}^{18}}}{2}\] \[=4.68\times {{10}^{18}}\]
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