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JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    The electron affinity of chlorine is 3.7 eV. 1 gram of chlorine is completely converted to \[C{{l}^{-1}}\] ion in a gaseous state. (1 eV = 23.06 kcal \[mo{{l}^{-1}}\]).Energy released in the process is   JEE Main Online Paper (Held On 07 May 2012)

    A) 4.8kcal                 

    B)                        7.2kcal

    C)                        8.2 kcal                

    D)                        2.4 kcal

    Correct Answer: D

    Solution :

                    Number of moles \[=\frac{1}{35.5}\] Given, 1 eV = 23.06 kcal \[\text{mo}{{\text{l}}^{\text{-1}}}\] eV = 3.7 x 23.06 kcal \[\text{mo}{{\text{l}}^{\text{-1}}}\] i.e. 1 mole realease energy \[=3.7\times 23.06\] kcal \[\therefore \]Energy released . \[=\frac{1}{35.5}\times 3.7\times 23.06\,\text{kcal}=2.4\,\text{kcal}\]


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