(A) Molarity $(M)$ of $BaCl_2 = \frac{\text{mass} \times 1000}{\text{Molar mass} \times \text{Volume in } mL} = \frac{1.0 \times 1000}{208 \times 200}

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(A) Molarity $(M)$ of $BaCl_2 = \frac{\text{mass} \times 1000}{\text{Molar mass} \times \text{Volume in } mL} = \frac{1.0 \times 1000}{208 \times 200} = 0.02404 \ M \approx 0.024 \ M$. Normality $(N)$ of $BaCl_2 = M \times \text{valency factor} = 0.024 \times 2 = 0.048 \ N$. Molar conductivity $(\Lambda_m)$ = $\frac{\kappa \times 1000}{M} = \frac{0.0058 \times 1000}{0.024} = 241.67 \ S \ cm^2 \ mol^{-1}$. Equivalent conductivity $(\Lambda_{eq})$ = $\frac{\kappa \times 1000}{N} = \frac{0.0058 \times 1000}{0.048} = 120.83 \ S \ cm^2 \ eq^{-1}$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Difficult

The specific conductance of a solution containing $1.0 \ g$ of anhydrous $BaCl_2$ in $200 \ cm^3$ of solution is $0.0058 \ S \ cm^{-1}$. Calculate the molar and equivalent conductivity of the solution. (Molar mass of $BaCl_2 = 208 \ g \ mol^{-1}$)

A
$241.67, \ 120.83$
B
$325.67, \ 250.83$
C
$193.67, \ 183.83$
D
$271.67, \ 130.83$

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Conductor and Conductance and Cell constant

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The specific conductance of a solution containing $1.0 \ g$ of anhydrous $BaCl_2$ in $200 \ cm^3$ of solution is $0.0058 \ S \ cm^{-1}$. Calculate the molar and equivalent conductivity of the solution. (Molar mass of $BaCl_2 = 208 \ g \ mol^{-1}$)

Similar Questions

The dissociation constant of $n$-butyric acid is $1.6 \times 10^{-5}$ and the molar conductivity at infinite dilution is $380 \times 10^{-4} \ S \ m^2 \ mol^{-1}$. The specific conductance of the $0.01 \ M$ acid solution is

When is the conductance $G$ equal to the conductivity $k$?

For an $HCl$ solution at $25^o C$,the equivalent conductance at infinite dilution is $425 \ \Omega^{-1} \ cm^2 \ eq^{-1}$. The electrical conductivity of the $HCl$ solution is $3.825 \ \Omega^{-1} \ cm^{-1}$. If the degree of dissociation is $90\%$,what is the normality of the solution in $N$?

The resistance of a $0.1 \, N$ solution of a salt is $2.5 \times 10^{3} \, \Omega$. If the cell constant is $1.15 \, cm^{-1}$,what will be the equivalent conductance of the solution in $\Omega^{-1} \, cm^{2} \, eq^{-1}$ (in $.6$)?

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