The molar conductivities of NaOH, NaCl and BaCl2 at infinite dilution are 2.481 × 10-2 S m2 mol-1, 1.265 × 10-2 S m2 mol-1 and 2.800 × 10-2 S m2 mol-1 respectively. The molar conductivity of Ba(OH)2 at infinite dilution will be

Question

The molar conductivities of NaOH, NaCl and BaCl2 at infinite dilution are 2.481 × 10-2 S m2 mol-1, 1.265 × 10-2 S m2 mol-1 and 2.800 × 10-2 S m2 mol-1 respectively. The molar conductivity of Ba(OH)2 at infinite dilution will be (A) 5.232 × 10-2 S m2 mol-1 (B) 9.654 × 10-2 S m2 mol-1 (C) 4.016 × 10-2 S m2 mol-1 (D) 1.145 × 10-2 S m2 mol-1

Tardigrade Admin · Accepted Answer

Lambda °m [Ba(OH)2] = Lambda °m (BaCl2) + 2 Lambda °m (NaOH) - 2 Lambda °m (NaCl) = (2.8 + 2 × 2.481 - 2 × 1.265) × 10-2 S m2 mo1-1 = (2.8 + 4. 962 - 2.53) × 10-2 S m2 mol-1 = 5.232 × 10-2 S m2 mol-1

$5.232 \times 1 0^{- 2} S m^{2} m o l^{- 1}$

$9.654 \times 1 0^{- 2} S m^{2} m o l^{- 1}$

$4.016 \times 1 0^{- 2} S m^{2} m o l^{- 1}$

$1.145 \times 1 0^{- 2} S m^{2} m o l^{- 1}$

Q. The molar conductivities of NaOH,NaCl and BaCl2​ at infinite dilution are 2.481×10−2Sm2mol−1,1.265×10−2Sm2mol−1 and 2.800×10−2Sm2mol−1 respectively. The molar conductivity of Ba(OH)2​ at infinite dilution will be

5.232×10−2Sm2mol−1

9.654×10−2Sm2mol−1

4.016×10−2Sm2mol−1

1.145×10−2Sm2mol−1

Λm∘​[Ba(OH)2] =Λm∘​(BaCl2​)+2Λm∘​(NaOH)−2Λm∘​(NaCl) = (2.8+2×2.481−2×1.265)×10−2Sm2mo1−1 = (2.8+4.962−2.53)×10−2Sm2mol−1 = 5.232×10−2Sm2mol−1

$5.232 \times 1 0^{- 2} S m^{2} m o l^{- 1}$

$9.654 \times 1 0^{- 2} S m^{2} m o l^{- 1}$

$4.016 \times 1 0^{- 2} S m^{2} m o l^{- 1}$

$1.145 \times 1 0^{- 2} S m^{2} m o l^{- 1}$