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BHU PMT BHU PMT Solved Paper-2003

  • question_answer
    How much chlorine will be liberated on passing one ampere current for 30 min through\[NaCl\]solution?

    A)  0.66 mol                             

    B)  0.33 mol

    C)  0.66g                                   

    D)  0.33 g

    Correct Answer: C

    Solution :

                     Key Idea: We will use Faraday's law to solve the problem \[\frac{w}{E}=\frac{it}{96500}\] Or           \[w=\frac{Eit}{96500}\] Given \[E=35.5,\text{ }i=1A,\] \[t=30\text{ }min=3\times 60\text{ }s\] Substituting values                 \[w=\frac{35.5\times 1\times 30\times 60}{96500}=0.66g\] \[\therefore \]0.66 g of chlorine is liberated.


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