Question

5. Conductivity of a solution containing one gram of
anhydrous BaCl2 in 200 ml of the solution has been found
to be 0.00585 ohm cm-1. Calculate the equivalent as
well as molar conductance,​

Answer 1

Answer:

Mass of Becl2=1 gram

Mass of solution=200mL

conductivity(k)=0.00587ohmcm

Molarity=\frac{Moles of solute}{Volume of solution in L}Molarity=

VolumeofsolutioninL

Molesofsolute

Moles of Becl2=\frac{Mass}{Molar mass}MolesofBecl2=

Molarmass

Mass

=\frac{1g}{79.9g/mol}

79.9g/mol

1g

=0.0125mol

Then the molarity M=\frac{0.0125mol}{0.2L}

0.2L

0.0125mol

=0.0625mol/L

And the equivalent conductivity(∧)=K×\frac{1000}{N}

N

1000

Here N=Normality=Molarity×gram equivalents

=0.625M×2

=1.25

=0.00587×\frac{1000}{1.25}

1.25

1000

=4.68ohm^{-1} cm^{3}4.68ohm

−1

cm

3

Molar conductivity(μ)=\frac{k*1000}{M}

M

k∗1000

=0.00587×\frac{1000}{0.0625}

0.0625

1000

=93.6cm^{3} ohm^{-1} mol^{-1}93.6cm

3

ohm

−1

mol

−1