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CET Karnataka Engineering CET - Karnataka Engineering 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 :

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


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