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BCECE Engineering BCECE Engineering Solved Paper-2014

  • question_answer
    19 g of a mixture containing \[\text{NaHC}{{\text{O}}_{\text{3}}}\]and \[N{{a}_{2}}C{{O}_{3}}\]on complete heating liberated 1.12 L of \[C{{O}_{2}}\] at STP. The weight of the remaining solid was 15.9 g. What is the weight (in g) of \[N{{a}_{2}}C{{O}_{3}}\]in the mixture before heating?

    A) 8.4                                         

    B) 15.9

    C) 4.0                                         

    D) 1.6

    Correct Answer: D

    Solution :

    Molecular weight of \[NaHC{{O}_{3}}=23+1+12+48=84\] Molecular weight of \[N{{a}_{2}}C{{O}_{3}}=46+12+48=106\] Hence, total weight \[=84+106=190\] \[\because \]In 190 g of a mixture, weight of \[NaC{{O}_{3}}\] is =106 \[\therefore \]  In 19 g of a mixture weight of \[N{{a}_{2}}C{{O}_{3}}=\frac{106\times 19}{190}\] \[=10.6g\]


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