Q

# 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)

$1:0.2$

Views

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