• question_answer If 1A of current is passed through $CuS{{O}_{4}}$solution for 10 s, the number of copper atoms deposite at the cathode will be about A)  $1.6\times {{10}^{20}}$                             B)  $8\times {{10}^{19}}$ C)  $3.1\times {{10}^{19}}$                             D)  $6.2\times {{10}^{19}}$

Number of Cu ions liberated is $n=\frac{I.\,t}{e\,\text{(valence}\,\text{electron)}}$ $n=\frac{1\times 10}{1.6\times {{10}^{-19}}\times 2}$ $n=3.125\times {{10}^{19}}$ $n\approx 3.1\times {{10}^{19}}$