Search any question & find its solution
Question:
Answered & Verified by Expert
The molar conductivities of $\mathrm{HCl}, \mathrm{NaCl}, \mathrm{CH}_{3} \mathrm{COOH}$, and $\mathrm{CH}_{3} \mathrm{COONa}$ at infinite dilution follow the order
Options:
Solution:
1099 Upvotes
Verified Answer
The correct answer is:
$\mathrm{HCl}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{NaCl}>\mathrm{CH}_{3} \mathrm{COONa}$
$\lambda_{\infty} \quad \mathrm{CH}_{3} \mathrm{COO}^{-}=40.9 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{Na}^{+}=50.10 \times 10^{-4} \quad \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{Cl}^{-} \quad=\quad 76.35 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{H}^{+}=349 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{x} \mathrm{HCl}=\lambda_{u^{t}}+\lambda_{\mathrm{c}^{-}} / \lambda_{\mathrm{Cl}, \mathrm{coon}}=\lambda_{\mathrm{al}, \mathrm{coo}^{-}}+\lambda_{\mathrm{u}^{+}}$
$\lambda_{\mathrm{Ncl}}=\lambda_{\mathrm{N}}+\lambda_{\mathrm{c} \mathrm{r}^{-}} / \lambda_{\mathrm{CH}_{3} \mathrm{com} \mathrm{N}}=\lambda_{\mathrm{CH}_{3} \mathrm{coo}^{-}}+\lambda_{\mathrm{N}^{+}}$
$\lambda_{\text {HC }}=425.35 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\mathrm{cH}_{3} \mathrm{coon}}=389.9 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\text {Nacl }}=126.45 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\mathrm{Cl}, \mathrm{coON}}=91 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\infty} \quad \mathrm{Na}^{+}=50.10 \times 10^{-4} \quad \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{Cl}^{-} \quad=\quad 76.35 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{\infty} \quad \mathrm{H}^{+}=349 \times 10^{-4} \quad \mathrm{Sm}^{2} \quad \mathrm{~mol}^{-1}$
$\lambda_{x} \mathrm{HCl}=\lambda_{u^{t}}+\lambda_{\mathrm{c}^{-}} / \lambda_{\mathrm{Cl}, \mathrm{coon}}=\lambda_{\mathrm{al}, \mathrm{coo}^{-}}+\lambda_{\mathrm{u}^{+}}$
$\lambda_{\mathrm{Ncl}}=\lambda_{\mathrm{N}}+\lambda_{\mathrm{c} \mathrm{r}^{-}} / \lambda_{\mathrm{CH}_{3} \mathrm{com} \mathrm{N}}=\lambda_{\mathrm{CH}_{3} \mathrm{coo}^{-}}+\lambda_{\mathrm{N}^{+}}$
$\lambda_{\text {HC }}=425.35 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\mathrm{cH}_{3} \mathrm{coon}}=389.9 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\text {Nacl }}=126.45 \mathrm{sm}^{2} / \mathrm{mol}$
$\lambda_{\mathrm{Cl}, \mathrm{coON}}=91 \mathrm{sm}^{2} / \mathrm{mol}$
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.