Search any question & find its solution
Question:
Answered & Verified by Expert
$31 \mathrm{~g}$ of ethylene glycol $\left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right)$ is dissolved in $600 \mathrm{~g}$ of water. The freezing point depression of the solution is $\left(K_f\right.$ for water is $\left.1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
Options:
Solution:
2833 Upvotes
Verified Answer
The correct answer is:
$1.55 \mathrm{~K}$
Given,
$W_B($ mass of ethylene glycol $)=31 \mathrm{~g}$
$W_A$ (mass of water $)=600 \mathrm{~g}$
$K_f($ for water $)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
and $M_B\left(\right.$ for $\left.\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right)=62$
$$
\begin{array}{ll}
\because & \Delta T_f=K_f \cdot \frac{W_B}{M_B} \times \frac{1000}{W_A} \\
\therefore & \Delta T_f=\frac{1.86 \times 31 \times 1000}{62 \times 600} \\
& =1.55 \mathrm{~K}
\end{array}
$$
$W_B($ mass of ethylene glycol $)=31 \mathrm{~g}$
$W_A$ (mass of water $)=600 \mathrm{~g}$
$K_f($ for water $)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
and $M_B\left(\right.$ for $\left.\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right)=62$
$$
\begin{array}{ll}
\because & \Delta T_f=K_f \cdot \frac{W_B}{M_B} \times \frac{1000}{W_A} \\
\therefore & \Delta T_f=\frac{1.86 \times 31 \times 1000}{62 \times 600} \\
& =1.55 \mathrm{~K}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.