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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • question_answer
    The coolant usually contains a solution of antifreeze prepared by mixing equal volumes of ethylene giycol, \[{{C}_{2}}{{H}_{4}}{{(OH)}_{2}}\] and water. The density of ethylene glycol is\[1.113\,\,g\,\,c{{m}^{-3}}\]. The freezing point of the mixture is (\[{{K}_{f}}\] for water \[=1.86\,kg\,\,mo{{l}^{-1}}\,{{K}^{-1}}\])

    A) \[-66.78{}^\circ C\]

    B) \[-33.39{}^\circ C\]

    C) \[-15.58{}^\circ C\]

    D) \[-1.86{}^\circ C\]

    Correct Answer: B

    Solution :

     Mixture contains equal volumes of ethylene glycol and water. 1 mL water = 1 water (\[{{w}_{2}}\], solvent) 1 mL ethylene glycol = 1.13 g ethylene glycol (\[{{w}_{1}}\] , solute) Depressing in freezing point \[\Delta {{T}_{f}}=\frac{1000{{K}_{f}}\times {{w}_{1}}}{{{m}_{1}}{{w}_{2}}}\] \[=\frac{1000\times 1.86\times 1.113}{62\times 1}\times 1={{33.39}^{o}}C\] \[\therefore \] Freezing point of mixture \[=0-{{33.39}^{o}}C\]


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