RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    The molarity of a solution containing \[5\,\,g\] of \[NaOH\] in \[450\,\,mL\] solution will be

    A) \[0.189\,\,mol\,\,d{{m}^{-3}}\] 

    B) \[0.278\,\,mol\,\,d{{m}^{-3}}\]

    C) \[0.556\,\,mol\,\,d{{m}^{-3}}\]

    D)        \[0.027\,\,mol\,\,d{{m}^{-3}}\]

    Correct Answer: B

    Solution :

    Molarity of the solution, if \[w\] gram of the solute is present in V mL of the solution           \[M=\frac{w}{Molar\,\,mass\,\,of\,\,solute}\times \frac{1000}{V}\]                 \[=\frac{5}{40}\times \frac{1000}{450}\]                 \[=0.278\,\,mol\,\,{{L}^{-1}}\]                 \[=0.278\,\,mol\,\,d{{m}^{-3}}\]


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