JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    The ammonia \[(N{{H}_{3}})\]released on quantitative reaction of 0.6 g urea \[(N{{H}_{2}}CON{{H}_{2}})\]with sodium hydroxide (NaOH) can be neutralized by [JEE MAIN Held on 07-01-2020 Evening]

    A) 200 ml of 0.4 N HCl

    B) 100 ml of 0.1 N HCl

    C) 200 ml of 0.2 N HCl

    D) 100 ml of 0.2 N HCl

    Correct Answer: D

    Solution :

    [d]             2 moles of \[N{{H}_{3}}\]will react with 2 mole of HCl. 0.6 g of urea give \[=\frac{0.6}{60}\times 2\,=0.02\]mol of \[N{{H}_{3}}\] \[100\times 0.2\text{ }N\text{ }HCl=0.02\text{ }mol\] of HCl

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