2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2013

  • question_answer
    100 mL of \[\text{N}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]is mixed with 100 mL of \[\text{1 M NaOH}\]solution. The resulting solution will be

    A)  highly acidic         

    B)  neutral

    C)  highly basic         

    D)  slightly acidic

    Correct Answer: B

    Solution :

     We know that \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] So, \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}},{{H}_{2}}S{{O}_{4}}\,NaOH\] \[1\times 100\] \[{{N}_{2}}\times 100\] \[{{N}_{2}}=1\] \[\because \]100 mL of \[\text{1 N }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]will neutralize 100 mL of \[\text{1N NaOH}\]so, the solution will be neutral.


You need to login to perform this action.
You will be redirected in 3 sec spinner