BHU PMT BHU PMT (Screening) Solved Paper-2009

  • question_answer
    The molarity of 98%\[{{H}_{2}}S{{O}_{4}}(d=1.8\text{ }g/mL)\]by weight, is

    A)  6 M                                      

    B)  18 M

    C)  10 M                                    

    D)  4 M

    Correct Answer: B

    Solution :

                     98%\[{{H}_{2}}S{{O}_{4}}\]means 98 g\[{{H}_{2}}S{{O}_{4}}\]in 100 g solution ie, volume of the solution \[=\frac{100}{1.8}mL\] \[=55.55\text{ }mL\] 98 g\[{{H}_{2}}S{{O}_{4}}=1\text{ }mol\] \[\therefore \] \[Molanty=\frac{Moles\text{ }of\text{ }the\text{ }solute}{volume\text{ }of\text{ }solution\text{ }(in\text{ }mL)}\times 1000\]                 \[=\frac{1}{55.55}\times 1000=18\,M\]


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