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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    A mixture of \[CaC{{l}_{2}}\] and \[NaCl\] weighing 4.44 g 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 0.56 g of CaO. The percentage of \[NaCl\] in the mixture (atomic mass of \[Ca=40\]) is

    A) 75                                          

    B) 30.6

    C) 25                                          

    D) 69.4

    Correct Answer: A

    Solution :

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


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