Question

The equivalent conductance of a 0.014 n solution of chloroacetic acid

Answer 1

Answer:

At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 Ω-1 cm2 mol-1 ... (b) 0.1 M chloro acetic acid

Answer 2

The equivalent conductance of a 0.014 n solution of chloroacetic acid is 1.13 x  $10^{-3}$ mho-$cm^2$/equivalent.

We must measure the conductance of the solution and divide it by the concentration of ions in the solution to determine the equivalent conductance of the solution.

In mho $cm^2$/equivalent units, the analogous conductance is stated.

The following formula may be used to get the equivalent conductance of a chloroacetic acid solution at 0.014 n:

1. The conductance of the solution (in mho) multiplied by the cell constant (in) gives the molar conductivity of the solution, which must first be determined.

2. The specific conductivity of the solution is then calculated by dividing the molar conductivity by the solution's concentration (in moles per litre).

3. The equivalent conductance of the solution is then calculated by dividing the specific conductivity by the equivalent weight of the solute.

The molecular weight divided by the quantity of acidic hydrogens yields the equivalent weight of chloroacetic acid.

Chloroacetic acid has one acidic hydrogen and a molecular mass of 94.5 g/mol.

Hence, 94.5/1 = 94.5 g/equivalent is the equivalent weight.

A conductivity metre may be used to experimentally determine the solution's specific conductivity.

Let's suppose that a measurement of the specific conductivity shows it to be 0.0015 mho/cm.

The solution has a concentration of 0.014 n, which indicates that each litre of the solution contains 0.014 moles of chloroacetic acid.

Typically, the manufacturer will supply the cell constant for the conductivity cell being used to measure the conductance of the solution. Consider using a cell constant of 1 .

The solution's molar conductivity is then:

Molar conductivity = Specific conductivity x cell constant

Molar conductivity = 0.0015 mho/cm x 1  $cm^{-1}$

Molar conductivity = 0.0015 mho

Specific conductivity of the solution is:

Specific conductivity = Molar conductivity / concentration

Specific conductivity = 0.0015 mho / 0.014 mol/L

Specific conductivity = 0.107 mho-$cm^2$/mol

Finally, the equivalent conductance of the solution is:

Equivalent conductance = Specific conductivity / equivalent weight

Equivalent conductance = 0.107 mho-$cm^2$/mol / 94.5 g/equivalent

Equivalent conductance = 1.13 x $10^{-3}$ mho-$cm^2$/equivalent

Hence, the equivalent conductance of a 0.014 n solution of chloroacetic acid is 1.13 x  $10^{-3}$ mho-$cm^2$/equivalent.

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