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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    A solution of\[Ni{{(N{{O}_{3}})}_{2}}\]is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode? [JEE Main 9-4-2019 Afternoon]

    A) 0.20                

    B) 0.05

    C) 0.10                

    D) 0.15

    Correct Answer: B

    Solution :

    0.1 eq. of \[N{{i}^{+2}}\] will be discharged. No. of eq = (No of moles) × (n-factor) 0.1= (No. of moles) × 2 No. of moles of \[Ni=\frac{0.1}{2}=0.05\]


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