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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    1.7 gm of an ammonium salt were treated with 100 ml of normal \[NaOH\] solution and boiled till no more of ammonia gas was given off. The excess of \[NaOH\] solution left over required 60 ml of normal sulphuric acid. The % of ammonia in salt is-

    A)  20%                                      

    B)  30%

    C)  40%                                      

    D)  50%

    Correct Answer: C

    Solution :

    Equivalent of NaOH left = Equivalent of acid \[1\times {{V}_{ml}}=60\times 1\] \[\therefore \]\[{{V}_{ml}}=60ml\] \[{{({{V}_{NaOH}})}_{used}}=100-60=40\,ml\] \[N{{H}_{4}}^{+}+O{{H}^{-}}\xrightarrow[{}]{{}}N{{H}_{3}}+{{H}_{2}}O\] No. of equivalents of NaOH= No. of equivalent of \[NH_{4}^{+}\]salt\[40\times 1\times {{10}^{-3}}=\]No. of eq. of \[NH_{4}^{+}\] Wt. of ammonia\[=40\times {{10}^{-3}}\times 17\]% of ammonia in salt\[=\frac{40\times {{10}^{-3}}\times 17}{1.7}\times 100=40%\]


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