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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    The amount of sugar \[\left( {{C}_{12}}{{H}_{22}}{{O}_{11}} \right)\] required to prepare 2L of its 0.1 M aqueous solution is: [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) 17.1 g                          

    B) 34.2 g

    C) 68.4 g  

    D)                  136.8 g

    Correct Answer: C

    Solution :

    \[{{\eta }_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=0.1\times 2=0.2\] \[{{\omega }_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=0.2\times 342=68.4\,gm\]


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