Question

conductivity of a solution containing 1 gram of anhydrous bacl2 in centimetre cube what has been found to be 0.0058 7 centimetre inverse what are the molar conductance and equivalent conductance of the solution (if weight of Ba =137 and cl=35.5​

Answer 1

Answer:

Mass of Becl2=1 gram

Mass of solution=200mL

conductivity(k)=0.00587ohmcm

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

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

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

=0.0125mol

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

=0.0625mol/L

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

Here N=Normality=Molarity×gram equivalents

=0.625M×2

=1.25

=0.00587×$\frac{1000}{1.25}$

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

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

=0.00587×$\frac{1000}{0.0625}$

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

Answer 2

Answer:

Explanation:

Mass of Bacl2 = 1 gram  (Given)

Let the  mass of solution = 200mL

Conductivity of the solution (k) = 0.00587 cm (Given)

Molarity = Moles of salute/ Volume of solution

Molar mass of BaCl2 = 137 + 2 ×35.5 = 208 g mol-1

Number of moles of BaCl2 in 200 cm3 of solution = 1/208

Volume of solution that contains 1 mol of BaCl2 (V) = 200 × 208cm³

Molar conductivity = Am × kV

= Am = 0.0058 × 200 × 208  

= 241.28 cm² mol-1

Therefore, the molar conductance of the element is 241.28 cm² mol-1.