Search any question & find its solution
Question:
Answered & Verified by Expert
Assuming the compounds to be completely dissociated in aqueous solution, identify the pair of the solutions that can be expected to be isotonic at the same temperature.
Options:
Solution:
1435 Upvotes
Verified Answer
The correct answer is:
0.03 M $\mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$
$\because$ For Isotonic solution
$$
\pi_{1}=\pi_{2} \text { i.e. } i_{1} C_{1} R T=i f_{2} R T
$$
For option $(c)$
$$
\begin{array}{l}
i_{1}=2, i_{2}=3 \\
C_{1}=0.03 \\
C_{2}=0.02
\end{array}
$$
Thus, $\quad i_{1} \times C_{1}=i_{2} \times C_{2}$
Hence, are isotonic, i.e.
$$
0.06=0.06
$$
In all other options, $i_{1} \times C_{1} \neq i_{2} \times C_{2}$ Thus, are not isotonic.
$$
\pi_{1}=\pi_{2} \text { i.e. } i_{1} C_{1} R T=i f_{2} R T
$$
For option $(c)$
$$
\begin{array}{l}
i_{1}=2, i_{2}=3 \\
C_{1}=0.03 \\
C_{2}=0.02
\end{array}
$$
Thus, $\quad i_{1} \times C_{1}=i_{2} \times C_{2}$
Hence, are isotonic, i.e.
$$
0.06=0.06
$$
In all other options, $i_{1} \times C_{1} \neq i_{2} \times C_{2}$ Thus, are not isotonic.
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.