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JEE Main & Advanced Sample Paper JEE Main Sample Paper-18

  • question_answer
    Suppose 5 gm of\[C{{H}_{3}}COOH\]is dissolved in one litre of ethanol. Assume that no reaction between them takes place. Calculate molality of resulting solution if density of ethanol is 0.789 gm/ml.

    A)  0.0856                 

    B)  0.0956

    C)  0.1056                                 

    D)  0.1156

    Correct Answer: C

    Solution :

     Weight of\[C{{H}_{3}}COOH\]dissolved = 5 g Eq. of\[C{{H}_{3}}COOH\]dissolved\[=\frac{5}{60}\] Volume of ethanol = 1 litre = 1000 ml. \[\therefore \] Weight of ethanol = 1000\[\times \]0.789 = 789 g \[\therefore \] Molality of solution\[=\frac{\text{Moles}\,\text{of}\,\text{solute}}{\text{Weight}\,\text{of}\,\text{solvent}\,\text{in}}\] \[=\frac{\frac{5}{60\times 789}}{1000}=0.1056\]


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