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NEET Sample Paper NEET Sample Test Paper-73

  • question_answer
    0.5 gm of fuming \[{{H}_{2}}S{{O}_{4}}\] (oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. The percentage of free 803 in the sample is

    A) \[30.6\,%\]                    

    B) \[40.6\,%\]

    C) \[20.6\,%\]                    

    D) \[50\,%\]

    Correct Answer: C

    Solution :

    \[Meq.\,\,of\,\,{{H}_{2}}S{{O}_{4}}+Meq.of\,\,S{{O}_{3}}=Meq.of\,\,NaOH\] \[\therefore \,\,\,\,\,\,\,\frac{\left( 0.5-x \right)}{98/2}\times 1000+\frac{x}{80/2}\times 1000=26.7\times \,0.4\]\[\therefore ~~x\,\,=\,\,0.103\] \[\therefore ~~%\, of \,S{{O}_{3}} = \frac{0.103}{0.5} \times  100 = 20.6\,%\]


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