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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

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

    A)  \[75\]             

    B)  \[30.6\]

    C)  \[25\]             

    D)  \[69.4\]

    Correct Answer: A

    Solution :

    \[\underset{1mol}{\mathop{CaC{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{1mol}{\mathop{CaO}}\,+C{{O}_{2}}\] \[\underset{1mol}{\mathop{CaC{{l}_{2}}}}\,+N{{a}_{2}}C{{O}_{3}}\xrightarrow{{}}\underset{1mol}{\mathop{CaC{{O}_{3}}}}\,+2Na\] \[1mol\,CaO\cong 1mol\,\,CaC{{l}_{2}}\] \[\frac{0.56}{56}mol\,CaO\cong 0.01mol\,\,CaC{{l}_{2}}\] \[=0.01\times 111g\,\,CaC{{l}_{2}}\] Thus, in the mixture, weight of \[NaCl=4.44-1.11=3.33g\] \[\therefore \] Percentage of \[NaCl=\frac{3.33}{4.44}\times 100\] \[=75%\]


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