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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2014

  • question_answer
    Volume occupied by single \[CsCl\] ion pair in a crystal is \[7.014\times {{10}^{-23}}\text{ }c{{m}^{3}}\]. The smallest \[Cs-Cs\]internuclear distance is equal to length of the side of the cube corresponding to volume of one \[CsCl\] ion pair. The smallest \[Cs\] to \[Cs\] internuclear distance is nearly

    A)  \[4.4\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B)  \[4.3\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)  \[4\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D)  \[4.5\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

     : Volume of \[CsCl={{a}^{3}}\] \[7.014\times {{10}^{-23}}c{{m}^{3}}={{a}^{3}}\] (a = edge length) \[a=\sqrt[3]{7.014\times {{10}^{-23}}}\] \[a=4.12\times {{10}^{-8}}cm=4.12\overset{\text{o}}{\mathop{\text{A}}}\,\approx 4\overset{\text{o}}{\mathop{\text{A}}}\,\]


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