The specific conductivity of a solution containing 1.0g of anhydrous BaCl2 in 200 cm3 of the solution has been found to be 0.0058 S cm-1. Calculate the molar conductivity of the solution. (Molecular wt. of BaCl2 = 208).

Question

The specific conductivity of a solution containing 1.0g of anhydrous BaCl2 in 200 cm3 of the solution has been found to be 0.0058 S cm-1. Calculate the molar conductivity of the solution. (Molecular wt. of BaCl2 = 208). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Molarity of BaCl2 = (1× 1000/208 × 200) = 0.024 M Also, Normality of BaCl2= 0.024 × 2 = 0.048 N ( ∵ N = M × Valency factor) Now, ∧m = k× (1000/CM) = (0.0058×1000/0.024) = 241.67 S cm2 mol-1

Q. The specific conductivity of a solution containing $1.0 g$ of anhydrous $B a C l_{2}$ in $200 c m^{3}$ of the solution has been found to be $0.0058 S c m^{- 1}$ . Calculate the molar conductivity of the solution. (Molecular wt. of $B a C l_{2} = 208)$ .

Molarity of $B a C l_{2} = \frac{1 \times 1000}{208 \times 200} = 0.024 M$
Also, Normality of $B a C l_{2} = 0.024 \times 2 = 0.048 N$
$(∵ N = M \times$ Valency factor)
Now, $\land_{m} = k \times \frac{1000}{C _{M}} = \frac{0.0058 \times 1000}{0.024}$
$= 241.67 S c m^{2} m o l^{- 1}$

Q. The specific conductivity of a solution containing 1.0g of anhydrous BaCl2​ in 200cm3 of the solution has been found to be 0.0058Scm−1. Calculate the molar conductivity of the solution. (Molecular wt. of BaCl2​=208).

Molarity of BaCl2​=208×2001×1000​=0.024M Also, Normality of BaCl2​=0.024×2=0.048N (∵N=M× Valency factor) Now, ∧m​=k×CM​1000​=0.0240.0058×1000​ =241.67Scm2mol−1

Q. The specific conductivity of a solution containing $1.0 g$ of anhydrous $B a C l_{2}$ in $200 c m^{3}$ of the solution has been found to be $0.0058 S c m^{- 1}$ . Calculate the molar conductivity of the solution. (Molecular wt. of $B a C l_{2} = 208)$ .

Molarity of $B a C l_{2} = \frac{1 \times 1000}{208 \times 200} = 0.024 M$
Also, Normality of $B a C l_{2} = 0.024 \times 2 = 0.048 N$
$(∵ N = M \times$ Valency factor)
Now, $\land_{m} = k \times \frac{1000}{C _{M}} = \frac{0.0058 \times 1000}{0.024}$
$= 241.67 S c m^{2} m o l^{- 1}$