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Engineering

How to solve this problem- - Solutions - JEE Main-2

1\; g of a non-volatile non-electrolyte solute is dissolved in 100 \; g of two different solvents A and B whose ebulliscopic constants are in the ratio of 1:5. The ratio of the elevation in their boiling points,\frac{\Delta T_{b}\left ( A \right )}{\Delta T_{b}\left ( B \right )}, is :

  • Option 1)

    5:1

  • Option 2)

    10:1

  • Option 3)

    1:5

  • Option 4)

     1:0.2

Answers (1)
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S solutionqc

\frac{\left ( \Delta T_{b} \right )_{A}}{\left ( \Delta T_{b} \right )_{B}}=?                                  \frac{\left ( K_{b} \right )_{A}}{\left (K_{b} \right )_{B}}=\frac{1}{5} given 

\left ( \Delta T_{b} \right )_{A}=\left ( K_{b} \right )_{A}\times m_{A}------------------------(1)

                                                                                 m_{A}=\frac{1}{m.w}\times \frac{1000}{100}

\left ( \Delta T_{b} \right )_{B}=\left ( K_{b} \right )_{B}\times m_{A}-------------------------(2)

     i\div ii

\frac{\left ( \Delta T_{b} \right )_{A}}{\left ( \Delta T_{b} \right )_{B}}=\frac{\left ( K_{b} \right )_{B}}{\left ( K_{b} \right )_{B}}\times \frac{m_{A}}{m_{B}}=\frac{1}{5}\times \frac{m_{A}}{m_{B}}

                 \frac{1}{5}\times \frac{\frac{1}{m.w}\times \frac{1000}{100}}{\frac{1}{m.w}\times \frac{1000}{100}}=\frac{1}{5}


Option 1)

5:1

Option 2)

10:1

Option 3)

1:5

Option 4)

 1:0.2

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