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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    A mixture of \[CaC{{l}_{2}}\] and \[NaCl\] weighing 4.44 g is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of \[CaO\]. The percentage of \[NaCl\] in the mixture is (atomic mass of\[Ca=40\])

    A)  \[75\]             

    B)  \[31.5\]

    C)  \[40.2\]            

    D)  \[25\]

    Correct Answer: A

    Solution :

    \[\underset{0.56g}{\mathop{CaO}}\,\xleftarrow{{}}\underset{1g}{\mathop{CaC{{O}_{3}}}}\,\xleftarrow{{}}\underset{1.12g}{\mathop{CaC{{l}_{2}}}}\,\] \[\left[ \begin{matrix}    56  \\    0.56  \\ \end{matrix}\to \begin{matrix}    100  \\    1g  \\ \end{matrix} \right]\] \[\left[ \begin{matrix}    100  \\    1g  \\ \end{matrix}\to \begin{matrix}    112  \\    1.12  \\ \end{matrix} \right]\] \[\therefore \] Amount of \[NaCl=4.44-1.12=3.32\] % of \[NaCl=\frac{3.32}{4.44}\times 100=75\]


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