JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    A \[5%\] solution (by mass) of cane sugar in water has freezing point of \[271\,K\]. The freezing point of \[5%\] glucose in water is :

    A)  \[269.2\,K\]     

    B) \[271K\]

    C) \[272\ K\]

    D) \[260\,K\]

    Correct Answer: A

    Solution :

    \[\Delta {{T}_{f}}={{K}_{f}}\times m\times i\] \[\frac{2}{\Delta {{T}_{f}}}=\frac{{{k}_{1}}\times \frac{5}{342}\times \frac{1000}{95}}{{{k}_{1}}\times \frac{5}{180}\times \frac{95}{1000}}\] \[\Delta {{T}_{f}}=\frac{342\times 2}{180}=3.8\] \[{{T}_{f}}=269.2K\]


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