Question

equal volumes of 0.015M CH3COOH and 0.015M NaOH are mixed together. what would be molar conductivity of CH3COONa is 6.3×10^-4 S/CM
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Answer 1

Given :

volume of  CH₃COOH = 0.015M

Volume of NaOH = 0.015M

Molar conductivity of CH₃COOH = 6.3×10⁻⁴ S/CM

To find :

Molar conductivity of the mixture

Solution :

• Molarity of CH₃COONa = 0.015V/2V

                                       =0.0075M

• Molar conductivity of mixture = K×1000/M

                                                 =6.3×10⁻⁴ ×1000 / 0.0075

                                                 =81M

• The Molar conductivity of the mixture is 81M

Answer 2

The molar conductivity of CH3COONa is 81 M

Given:

The volume of CH₃COOH = 0.015 M

The volume of NaOH = 0.015 M

The molar conductivity of CH₃COOH = 6.3×10⁻⁴ S/CM

To find:

Molar conductivity of CH₃COONa = ?

Step-by-step explanation:

CH₃COOH + NaOH → CH₃COONa

Molality of CH₃COONa = 0.015/2

∴ Molality of CH₃COONa = 0.0075 M

The molar conductivity is given by the formula:

λ = (K × 1000)/M

On substituting the values, we get,

λ = (6.3 × 10⁻⁴ × 1000)/0.0075

∴ λ = 81 M