Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The resistance of a 0.1 N solution of acetic acid is 250 $\Omega$ when measured in a cell of cell constant 1.15 cm^-1. The equivalent conductance (in ohm^-1 cm² eq^-1) of 0.1 N acetic acid is:
Option: 1
9.2

Option: 2
24.6

Option: 3
36.8

Option: 4
46

Answers (1)

best_answer

$\begin{array}{l}{\kappa=G \times G^{*}=\frac{1}{R} \times G^{*}=\frac{1}{250} \times 1.15} \\\\ {\Lambda_{\mathrm{eq}}=\frac{\kappa \times 1000}{N}=\frac{1.15 \times 1000}{250 \times 0.1}=46 \mathrm{\: S} \mathrm{\: cm}^{2} \mathrm{\: eq}^{-1}}\end{array}$

Therefore, option(4) is correct

Posted by

mansi

View full answer

Similar Questions

0.1 M acetic acid solution having pH =3 is titrated against 0.05 M NaOH solution. Calculate pH at 1/4th stage of neutralization of acid.Option: 1

0.25×60+0.10x=0.15×(60+x)

Trending Articles/News

anam-khan

TS EAMCET 2025 admit card out on eapcet.tgche.ac.in; exams from April 29

anam-khan

TS EAMCET 2025 admit card today at eapcet.tgche.ac.in

anam-khan

TS EAMCET 2025 application correction facility available till April 8

Latest Question