The equivalent conductance of a monobasic aci | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $1$. Calculate the normality $(N)$ of the solution: $N = \frac{	ext{mass}}{	ext{equivalent weight} 	imes 	ext{volume in L}} = \frac{15}{49 	i

Vedclass · Accepted Answer

$51.7$

Vedclass · Answer

(D) $1$. Calculate the normality $(N)$ of the solution: $N = \frac{	ext{mass}}{	ext{equivalent weight} 	imes 	ext{volume in L}} = \frac{15}{49 	imes 1} \approx 0.306 \ eq \ L^{-1}$.
$2$. Calculate the specific conductance $(\kappa)$: $\kappa = \frac{1}{	ext{resistivity}} = \frac{1}{18.5} \approx 0.05405 \ ohm^{-1} \ cm^{-1}$.
$3$. Calculate the equivalent conductance $(\Lambda_{eq})$ at the given concentration: $\Lambda_{eq} = \frac{\kappa 	imes 1000}{N} = \frac{0.05405 	imes 1000}{0.306} \approx 176.63 \ ohm^{-1} \ cm^2 \ eq^{-1}$.
$4$. Calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\Lambda_{eq}}{\Lambda_{\infty}} = \frac{176.63}{348} \approx 0.5075$.
$5$. Express as a percentage: $\alpha \% = 0.5075 	imes 100 \approx 50.75 \% \approx 51.7 \%$.

List-$I$ (Symbol of electrical property)	List-$II$ (Units)
$A.$ $\wedge_m$	$I.$ $S\,cm^2\,mol^{-1}$
$B.$ $G$	$II.$ $S$
$C.$ $\kappa$	$III.$ $S\,cm^{-1}$
$D.$ $G^*$	$IV.$ $cm^{-1}$

The equivalent conductance of a monobasic acid at infinite dilution is $348 \ ohm^{-1} \ cm^2 \ eq^{-1}$. If the resistivity of the solution containing $15 \ g$ of the acid (molecular weight $49$) in $1 \ L$ is $18.5 \ ohm \ cm$,what is the degree of dissociation of the acid? (Answer in percentage)

Similar Questions

Match the following:
List-$I$ (Symbol of electrical property) List-$II$ (Units)
$A.$ $\wedge_m$ $I.$ $S\,cm^2\,mol^{-1}$
$B.$ $G$ $II.$ $S$
$C.$ $\kappa$ $III.$ $S\,cm^{-1}$
$D.$ $G^*$ $IV.$ $cm^{-1}$

The correct answer is

The molar conductivities of $HCl$,$NaCl$,$CH_{3}COOH$,and $CH_{3}COONa$ at infinite dilution follow the order:

If the values of $\Lambda_{\infty}$ of $NH_4Cl$,$NaOH$ and $NaCl$ are $130$,$217$ and $109 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ respectively,the $\Lambda_{\infty}$ of $NH_4OH$ in $ohm^{-1} \ cm^2 \ equiv^{-1}$ is:

Molar conductivities $\left(\Lambda_{m}^{\circ}\right)$ at infinite dilution of $NaCl$,$HCl$,and $CH_{3}COONa$ are $126.4$,$425.9$,and $91.0 \ S \ cm^{2} \ mol^{-1}$ respectively. $\Lambda_{m}^{\circ}$ for $CH_{3}COOH$ will be $:-$

Which solution of $NaNO_3$ will have maximum specific conductance in $N$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module