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BCECE Medical BCECE Medical Solved Papers-2014

  • question_answer
    19 g of a mixture containing \[NaHC{{O}_{3}}\] and \[NaC{{O}_{3}}\] 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) 10.6

    Correct Answer: D

    Solution :

    Molecular weight of                 \[NaHC{{O}_{3}}=23+1+12+48=48\]                 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 \[N{{a}_{2}}C{{O}_{3}}\] 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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