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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    8 g of NaOH is dissolved in 18 g of \[{{H}_{2}}O.\] Mole fraction of \[NaOH\]in solution and molality (in \[mol\,k{{g}^{-1}}\]) of the solution respectively are [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) 0.167, 22.20                             

    B) 0.167, 11.11

    C) 0.2, 22.20        

    D)   0.2, 11.11

    Correct Answer: B

    Solution :

    Mole of \[NaOH=\frac{8}{40}=0.2\] Mole of water \[=\frac{18}{18}=1\] Total moles \[=1+0.2=1.2\] Mole fraction of \[NaOH\] \[=\frac{Moles\,of\,NaOH}{Total\,moles}=\frac{0.2}{1.2}=0.167\] \[Molality=\frac{Moles\,of\,NaOH}{Mass\,of\,solvent}\times 1000\] \[=\frac{0.2}{18}\times 1000=11.11m\]


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