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 conditions ⇒

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

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

Degree of the Dissociation of the HCl acid(α) = 90%
= 90/100 
= 0.9

Now, 
 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    [1 cm³ = 10⁻³ L]

Since, n factor for the HCl is 1.

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

Hence, the Normality of the solution is 1 N.


Hope it helps.