Search any question & find its solution
Question:
Answered & Verified by Expert
The molar conductances of $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{CH}_3 \mathrm{COONa}$ at infinite dilution are $126.45,426.16$ and $91 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The molar conductance of $\mathrm{CH}_3 \mathrm{COOH}$ at infinite dilution is
Options:
Solution:
1730 Upvotes
Verified Answer
The correct answer is:
$390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\wedge_m^o\left(\mathrm{CH}_3 \mathrm{COOH}\right)=$
$\wedge^0\left(\mathrm{CH}_3 \mathrm{COONa}\right)+\wedge^{\circ}(\mathrm{HCl})-\wedge^{\circ}(\mathrm{NaCl})$
$=91+426.16-126.45=390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\wedge^0\left(\mathrm{CH}_3 \mathrm{COONa}\right)+\wedge^{\circ}(\mathrm{HCl})-\wedge^{\circ}(\mathrm{NaCl})$
$=91+426.16-126.45=390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
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.