Molar conductivities at infinite dilution wed | vedclass

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

Question

(C) According to Kohlrausch's law of independent migration of ions: $\wedge_{m}^{\circ}(NH_4OH) = \wedge_{m}^{\circ}(NH_4^+) + \wedge_{m}^{\circ}(OH^{

Vedclass · Accepted Answer

$321$

Vedclass · Answer

(C) According to Kohlrausch's law of independent migration of ions: $\wedge_{m}^{\circ}(NH_4OH) = \wedge_{m}^{\circ}(NH_4^+) + \wedge_{m}^{\circ}(OH^{-})$ We can express this using the given values: $\wedge_{m}^{\circ}(NH_4OH) = \wedge_{m}^{\circ}(NH_4Cl) + \frac{1}{2} \wedge_{m}^{\circ}(Ba(OH)_2) - \frac{1}{2} \wedge_{m}^{\circ}(BaCl_2)$ Substituting the given values: $\wedge_{m}^{\circ}(NH_4OH) = 213.0 + (\frac{1}{2} 	imes 457.0) - (\frac{1}{2} 	imes 240.6)$ $\wedge_{m}^{\circ}(NH_4OH) = 213.0 + 228.5 - 120.3 = 321.2 \ S \ cm^2 \ mol^{-1}$

Molar conductivities at infinite dilution $\wedge_{m}^{\circ}$ for $Ba(OH)_2$,$BaCl_2$ and $NH_4Cl$ are $457.0$,$240.6$ and $213.0 \ S \ cm^2 \ mol^{-1}$ respectively. The $\wedge_{m}^{\circ}$ for ammonium hydroxide (in $S \ cm^2 \ mol^{-1}$) is (in $.2$)

Similar Questions

If the specific conductance of $1 \, M \, H_2SO_4$ solution is $26 \times 10^{-2} \, S \, cm^{-1}$,then the equivalent conductivity would be

$0.5 \ N$ solution of a salt placed between two platinum electrodes $2.0 \ cm$ apart and of area of cross section $4.0 \ cm^2$ has a resistance of $25 \ \Omega$. Calculate the equivalent conductivity of solution ................. $\Omega^{-1} \ cm^2 \ eq^{-1}$

Among the following,the plot that correctly represents the conductometric titration of $0.05 \ M \ H_2SO_4$ with $0.1 \ M \ NH_4OH$ is

What is the relation between the concentration of solution and molar conductivity $(\Lambda_m)$?

Which of the following decreases on dilution of an electrolytic solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module