BCECE Medical BCECE Medical Solved Papers-2004

  • question_answer
    A current of strength 2.5A was passed through\[CuS{{O}_{4}}\] solution for 6 min 26s. The amount of copper deposited is : (At. wt. of Cu = 63.5,1 F = 96500 C)

    A)  0.3175 g       

    B)  3.175 g

    C)  0.635 g        

    D)  6.35 g

    Correct Answer: A

    Solution :

    Given \[i=2.5\,A\]      \[t=6\,\min \,26\,s=6\times 60+26=386\,\,s\] No. of coulomb passed \[=i\times t\]                  \[=2.5\times 386\]                  \[=965\,\,C\]                 \[C{{u}^{2+}}+2e\xrightarrow{{}}Cu\] \[\therefore \] \[2\times 96500\,\,C\] charge deposits \[Cu=63.5\,g\] \[\therefore \] 965 C charge deposits                 \[Cu=\frac{63.5}{2\times 96500}\times 965\] \[=0.3175\,g\]


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