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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    A sample of \[NaCl{{O}_{3}}\] is converted by heat to \[NaCl\] with a loss of 0.16g of oxygen. The residue is dissolved in water and precipitated as \[AgCl\]. The mass of \[AgCl\] (in g) obtained will be: (Given: Molar mass of \[\text{AgCl=143}\text{.5g}\,\,\text{mo}{{\text{l}}^{\text{-1}}}\])                                                                                                                                           [JEE Online 15-04-2018]

    A) 0.41                

    B) 0.35    

    C) 0.48                

    D) 0.54

    Correct Answer: C

    Solution :

    The molar mass of \[{{\text{O}}_{2}}=32g/mol\] \[0.16g\] of oxygen \[=\frac{0.16g}{32g/mol}=0.005mol\] \[2NaCl{{O}_{3}}=2NaCl+3{{O}_{2}}\] 3 moles of \[{{O}_{2}}=2\] moles of \[NaCl=2\] moles of \[AgCl.\] 0.005 moles of \[{{O}_{2}}=0.005\times \frac{2}{3}=0.003333\] moles of \[AgCl\]. Molar mass of \[AgCl=143.5g\,\,mo{{l}^{-1}}\] The mass of \[AgCl\] (in g) obtained will be \[=143.5g\,\,mo{{l}^{-1}}\times 0.003333mol=0.48g.\]


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