J & K CET Medical J & K - CET Medical Solved Paper-2014

  • question_answer
    Molality of 2.5 g of ethanoic acid\[(C{{H}_{3}}COOH)\]in 75 g of benzene is

    A)  \[0.565\text{ }mol\text{ }k{{g}^{-1}}\]     

    B)  \[0.656\text{ }mol\text{ }k{{g}^{-1}}\]

    C)  \[0.556\text{ }mol\text{ }k{{g}^{-1}}\]     

    D)  \[0.665\text{ }mol\text{ }k{{g}^{-1}}\]

    Correct Answer: C

    Solution :

     Molality (m) \[=\frac{mass\text{ }of\text{ }solute\text{ (}ing)\times 1000}{molecular\text{ }weight\text{ }of\text{ }solute\times mass\text{ }of\text{ }solvent\text{ }(ing)}\] \[=\frac{2.5\times 1000}{60\times 75}\] \[=0.555mol\,k{{g}^{-1}}\]


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