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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    How many grams of methyl alcohol should added to 10 litre tank of water to prevent its freezing at 268 K?                 (\[{{\operatorname{K}}_{f}}\]for water is 1.86 K kg \[{{\operatorname{mol}}^{-1}}\])     JEE Main Online Paper ( Held On 25  April 2013 )

    A) 880.07 g                               

    B)                        899.04 g

    C)                        866.02 g                               

    D)                        868.06 g

    Correct Answer: D

    Solution :

                    \[\Delta {{T}_{f}}={{K}_{f}}m\]where m = molality \[273-268=1.86\times \frac{w}{M\times v}\] \[5=1.86\times \frac{w}{32\times 10}\] \[w=\frac{5\times 32\times 10}{1.86}\]                 \[=860.2\approx 868.06g\]


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