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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2014

  • question_answer
    \[10g\]of a mixture of \[BaO\]and \[CaO\]requires \[100c{{m}^{3}}\]of \[2.5M\text{ }HCl\]to react completely The percentage of calcium oxide in the mixture is approximately (Given: molar mass of\[BaO=153\])

    A)  \[52.6\]

    B)  \[55.1\]

    C)  \[44.9\]

    D)  \[47.4\]

    Correct Answer: A

    Solution :

    : \[BaO+2HCl\to BaC{{l}_{2}}+{{H}_{2}}O\]     \[CaO+2HCl\to CaC{{l}_{2}}+{{H}_{2}}O\] Reactions show that 2 moles of mix. of \[BaO\] and \[CaO\] requires 4 moles of \[HCl\] to react completely. Hence, no. of moles of \[HCl=\frac{2.5\times 100}{1000}\]      \[=0.25moles\]
    Now,   HC1    \[BaO+\text{C}aO\]
    Initial    4 moles        2 moles
    Final   \[0.25\]moles        ?
    No. of moles of mixture of \[BaO\] and \[\text{C}aO\]      \[=\frac{0.25\times 2}{4}=0.125moles\] Let the mass of \[CaO\] be xg and \[BaO\] is \[(10-x)g\] Then, \[\frac{x}{56}+\frac{10-x}{153}=0.125\]     [MoL mass of\[CaO=56,\text{ }BaO=153\]] \[153x+56(10-x)=0.125\times 56\times 153\] \[153x+560-56x=0.125\times 56\times 153\] \[97x=1071-560\] \[97x=511\] \[x=5.268\] \[\therefore \] % of \[CaO=\frac{5.268}{10}\times 100=52.68%\]


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