Equivalent conductivity of Cr_2(SO_4)_3 is re | vedclass

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

Question

(D) The relationship between molar conductivity $(\Lambda_m)$ and equivalent conductivity $(\Lambda_{eq})$ is given by the formula: $\Lambda_{eq} = \f

Vedclass · Accepted Answer

$\Lambda_{eq} = \frac{\Lambda_m}{6}$

Vedclass · Answer

(D) The relationship between molar conductivity $(\Lambda_m)$ and equivalent conductivity $(\Lambda_{eq})$ is given by the formula: $\Lambda_{eq} = \frac{\Lambda_m}{n-factor}$.For $Cr_2(SO_4)_3$,the dissociation is $Cr_2(SO_4)_3 ightarrow 2Cr^{3+} + 3SO_4^{2-}$.The total positive charge is $2 	imes (+3) = +6$ and the total negative charge is $3 	imes (-2) = -6$.Thus,the $n-factor$ (valency factor) is $6$.Therefore,$\Lambda_{eq} = \frac{\Lambda_m}{6}$.

Equivalent conductivity of $Cr_2(SO_4)_3$ is related to molar conductivity by the expression

Similar Questions

Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

If $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$, then $A+C-B$ is equal to :

At $25 \ ^\circ C,$ the solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-11}.$ At which $pH$ will $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001 \ M \ Mg^{2+}$ ions?

Black Jack is an ore of

Vedclass Test Series

Exam Paper Generator

Online Exam Module