Question

For hcl solution at 25°c, equivalent conductance at infinite dilution, is 425 ohm–1 cm2 equiv–1. the specific conductance of a solution of hcl is 0.3825 ohm–1 cm–1. if the apparent degree of dissociation is 90% the normality of the solution is

Answer 1

Hello Dear.

Given ⇒

Equivalent Conductance at Infinite(Λ infinite) = 425 Ω⁻¹cm² equi⁻¹.
Specific Conductance(Κ) = 0.3825 Ω⁻¹ cm⁻¹.

Degree of the Association(α) = 90%
= 90/100 
= 0.9

Now,We know,
α × Λ infinity = Λ at concentration c.
∴ Λ at concentration c = 0.9 × 425
∴ Λ at concentration c = 382.5

Now, C = Κ/Λ at concentration c.
∴ C = 0.3825/382.5
∴ C = 0.001 mol/cm³
∴ C = 1 mol/L

We know, n factor for the Hydrogen chloride is 1.

∴ Normality = C × n factor.
∴ Normality = 1 × 1
∴ Normality = 1 N.


Hence, the Normality is 1 N.


Hope it helps.

Answer 2

Answer:

Explanation:Hello Dear.

Given ⇒

Equivalent Conductance at Infinite(Λ infinite) = 425 Ω⁻¹cm² equi⁻¹.

Specific Conductance(Κ) = 0.3825 Ω⁻¹ cm⁻¹.

Degree of the Association(α) = 90%

= 90/100 

= 0.9

Now,We know,

α × Λ infinity = Λ at concentration c.

∴ Λ at concentration c = 0.9 × 425

∴ Λ at concentration c = 382.5

Now, C = Κ/Λ at concentration c.

∴ C = 0.3825/382.5

∴ C = 0.001 mol/cm³

∴ C = 1 mol/L

We know, n factor for the Hydrogen chloride is 1.

∴ Normality = C × n factor.

∴ Normality = 1 × 1

∴ Normality = 1 N.

Hence, the Normality is 1 N.

Hope it helps.

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