Question

The Equivalent conductivities of AgNO3, KBr, and KNO3 at infinite dilutions were found to be 120.0,138.0, and 132 ohm-1cm2eq-1 respectively. Find the equivalent conductance of AgBr at infinite dilution. If the specific conductivity of 0.001N solution of AgBr at 25oC is 3 X 10-5 ohm-1 cm-1. What is the degree of dissociation?​

Answer 1

Answer:

Sorry i guess its wrong answer

Explanation:

sorry.......................................

Answer 2

Answer:

The equivalent conductance of AgBr at infinite dilution is equal to 126 ohm⁻¹cm²eq⁻¹ and degree of dissociation is 23.8%.

Explanation:

Given, the equivalent conductance of AgNO₃, KBr, and KNO₃ at infinite dilutions were found to be 120.0, 138.0, and 132 ohm⁻¹cm²eq⁻¹ respectively.

AgNO₃  +   KBr  $\longrightarrow$  AgBr   +   KNO₃

The equivalent conductance of AgBr at infinite dilution:-

$\lambda^o(AgBr) = \lambda^o(AgNO_3)+\lambda^o(KBr)-\lambda^o(KNO_3)$

$\lambda^o(AgBr)$ = 120 + 138 - 132

$\lambda^o_{eq}(AgBr)$ = 126 ohm⁻¹cm²eq⁻¹

Given, concentration of solution ,C = 0.001N = 0.001 eq/L

Specific conductance , k = 3 ×10⁻⁵ohm⁻cm⁻¹

We know that equivalent conductance at concentration C,

$\lambda_{eq}=\frac{\kappa\times 1000}{C}$

$\lambda_{eq}=\frac{3\times 10^{-5}\times 1000}{0.001}$

$\lambda _{eq}=3\times 10^6\times10^{-5}$

$\lambda_{eq}=30\;ohm^{-1}cm^{2}eq^{-1}$

Degree of dissociation can be calculated by dividing the equivalent conductance at concentration C to equivalent conductance at infinite dilution.

$\alpha =\frac{\lambda_{eq}}{\lambda^o_{eq}}$

$\alpha =\frac{30}{126}$

$\alpha =0.238$ or $\alpha = 23.8\%$

Therefore, the equivalent conductance of AgBr at infinite dilution is equal to 126 ohm⁻¹cm²eq⁻¹  and degree of dissociation is 23.8%.

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