Search any question & find its solution
Question:
Answered & Verified by Expert
A 0.1 molal solution of an acid is $4.5 \%$ ionized. Calculate freezing point. (molecular weight of the acid is 300$) . K_f=1.86 \mathrm{~K} \mathrm{~mol}^{-1} \mathrm{~kg}$.
Options:
Solution:
1802 Upvotes
Verified Answer
The correct answer is:
$-0.194^{\circ} \mathrm{C}$
If acid is $4.5 \%$ ionized then $\alpha=0.045$.
$\begin{aligned} & \Delta T_f=\text { molality } \times K_f=0.1 \times 1.86=0.186 \\ & \Delta T_{\exp }=\Delta T_f(1+\alpha)=0.186(1+0.045)=0.194^{\circ} \mathrm{C}\end{aligned}$
$\begin{aligned} & \Delta T_f=\text { molality } \times K_f=0.1 \times 1.86=0.186 \\ & \Delta T_{\exp }=\Delta T_f(1+\alpha)=0.186(1+0.045)=0.194^{\circ} \mathrm{C}\end{aligned}$
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.