JEE MAIN - Chemistry (2022 - 27th June Morning Shift - No. 12)
2 g of a non-volatile non-electrolyte solute is dissolved in
200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio $${x \over y}$$ (x : y). The value of y is ______________. (Nearest integer)
Answer
8
Explanation
$\Delta T b=k b m$
$$ \begin{aligned} &\frac{\left(\Delta T_{b}\right)_{A}}{\left(\Delta T_{b}\right)_{B}}=\frac{\left(k_{b}\right)_{A}}{\left(k_{b}\right)_{B}} \\\\ &=\frac{1}{8}=\frac{x}{y} \\\\ &\therefore y=8 \end{aligned} $$
$$ \begin{aligned} &\frac{\left(\Delta T_{b}\right)_{A}}{\left(\Delta T_{b}\right)_{B}}=\frac{\left(k_{b}\right)_{A}}{\left(k_{b}\right)_{B}} \\\\ &=\frac{1}{8}=\frac{x}{y} \\\\ &\therefore y=8 \end{aligned} $$
Comments (0)
