Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • If the molar conductivity $\left(\Lambda_{\mathrm{m}}\right)$ of a

Q.46) If the molar conductivity $\left(\Lambda_{\mathrm{m}}\right)$ of a $0,050 \mathrm{~mol} \mathrm{~L}^{-1}$ solution of a monobasic weak acid is $90 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, its extent (degree) of dissociation will be [Assume $A_{+}^{\circ}=349.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and $$ \left.\mathrm{A}_{-}^{\mathrm{E}}=50.4 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\right] $$

A) 0.215

B) 0.115

C) 0.125

D) 0.225

 

Answers (1)

best_answer

To calculate the degree of dissociation ( $\alpha$ ) of the weak monobasic acid, we use the formula:

$$
\alpha=\frac{\Lambda_m}{\Lambda_m^{\circ}}
$$
Given that

$\begin{aligned} & \Lambda_m=90 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \Lambda_m^{\circ}=A_{+}^{\circ}+A_{-}^{\circ}=349.6+50.4=400 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$

so put the value in $$
\alpha=\frac{\Lambda_m}{\Lambda_m^{\circ}}
$$

$\alpha=\frac{90}{400}=0.225$

Hence the correct option is (4)

Posted by

Saumya Singh

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

CENTAC Puducherry to host interactive session for Non-NEET UG courses at 6 pm today
anam-khan

‘NTA failed to provide…’: MP High Court puts NEET result 2025 on hold over power failure at exam centre
anam-khan

NEET 2025 Live: NTA NEET answer key at neet.nta.nic.in soon; result date, MBBS cut-off updates

Latest Question