At 25° C, the molar conductances at infinite dilution for the strong electrolytes NaOH , NaCl and BaCl 2 are 248 × 10-4, 126 × 10-4 and 280 × 10-4 Sm 2 mol -1 respectively, λm° Ba ( OH )2 in Sm 2 mol -1 is

Question

At 25° C, the molar conductances at infinite dilution for the strong electrolytes NaOH , NaCl and BaCl 2 are 248 × 10-4, 126 × 10-4 and 280 × 10-4 Sm 2 mol -1 respectively, λm° Ba ( OH )2 in Sm 2 mol -1 is (A) 52.4 × 10-4 (B) 524 × 10-4 (C) 402 × 10-4 (D) 262 × 10-4

Tardigrade Admin · Accepted Answer

BaCl 2+2 NaOH arrow Ba ( OH )2+2 NaCl λm∞ Ba ( OH )2=λm∞ BaCl 2+2 λm∞ NaOH -2 λm∞ NaCl =280 × 10-4+2 × 248 × 10-4-2 × 126 × 10-4 =(280+496-252) × 10-4 =524 × 10-4 Sm 2 mol -1

Q. At $2 5^{\circ} C$ , the molar conductances at infinite dilution for the strong electrolytes $N a O H, N a Cl$ and $B a C l_{2}$ are $248 \times 1 0^{- 4}, 126 \times 1 0^{- 4}$ and $280 \times 1 0^{- 4} S m^{2} m o l^{- 1}$ respectively, $λ_{m}^{\circ} B a (O H)_{2}$ in $S m^{2} m o l^{- 1}$ is

$52.4 \times 1 0^{- 4}$

$524 \times 1 0^{- 4}$

$402 \times 1 0^{- 4}$

$262 \times 1 0^{- 4}$

Q. At 25∘C, the molar conductances at infinite dilution for the strong electrolytes NaOH,NaCl and BaCl2​ are 248×10−4,126×10−4 and 280×10−4Sm2mol−1 respectively, λm∘​Ba(OH)2​ in Sm2mol−1 is

52.4×10−4

524×10−4

402×10−4

262×10−4

BaCl2​+2NaOH→Ba(OH)2​+2NaCl λm∞​Ba(OH)2​=λm∞​BaCl2​+2λm∞​NaOH−2λm∞​NaCl =280×10−4+2×248×10−4−2×126×10−4 =(280+496−252)×10−4 =524×10−4Sm2mol−1

$52.4 \times 1 0^{- 4}$

$524 \times 1 0^{- 4}$

$402 \times 1 0^{- 4}$

$262 \times 1 0^{- 4}$