Molar conductivities ( Lambda m °) at infinite dilution of NaCl , HCl and CH 3 COONa are 126.4,425.9 and 91.0 S cm 2 mol -1 respectively. Lambda m ° for CH 3 COOH will be

Question

Molar conductivities ( Lambda m °) at infinite dilution of NaCl , HCl and CH 3 COONa are 126.4,425.9 and 91.0 S cm 2 mol -1 respectively. Lambda m ° for CH 3 COOH will be (A)  180.5 S cm2 mol-1  (B) 290.8 S cm2 mol-1  (C)  390.5 S cm2 mol-1  (D)  425.5 S cm2 mol-1

Tardigrade Admin · Accepted Answer

CH 3 COONa + HCl longrightarrow NaCl + CH 3 COOH Lambda m ( CH 3 COOH )°= Lambda m ( CH 3 COONa )°+ Lambda m ( HCl )°- Lambda m ( NaCl )° Lambda m ( CH 3 COOH )° =(91.0+425.9-126.4) S cm 2 mol -1 Lambda m ( CH 3 COOH )° =390.5 S cm 2 mol -1

Q. Molar conductivities $(Λ_{m}^{\circ})$ at infinite dilution of $N a Cl, H Cl$ and $C H_{3} COON a$ are $126.4, 425.9$ and $91.0 S c m^{2} m o l^{- 1}$ respectively. $Λ_{m}^{\circ}$ for $C H_{3} COO H$ will be

$180.5 S c m^{2} m o l^{- 1}$

$290.8 S c m^{2} m o l^{- 1}$

$390.5 S c m^{2} m o l^{- 1}$

$425.5 S c m^{2} m o l^{- 1}$

Q. Molar conductivities (Λm∘​) at infinite dilution of NaCl,HCl and CH3​COONa are 126.4,425.9 and 91.0Scm2mol−1 respectively. Λm∘​ for CH3​COOH will be

180.5Scm2mol−1

290.8Scm2mol−1

390.5Scm2mol−1

425.5Scm2mol−1

CH3​COONa+HCl⟶NaCl+CH3​COOH Λm(CH3​COOH)∘​=Λm(CH3​COONa)∘​+Λm(HCl)∘​−Λm(NaCl)∘​ Λm(CH3​COOH)∘​=(91.0+425.9−126.4)Scm2mol−1 Λm(CH3​COOH)∘​=390.5Scm2mol−1

Q. Molar conductivities $(Λ_{m}^{\circ})$ at infinite dilution of $N a Cl, H Cl$ and $C H_{3} COON a$ are $126.4, 425.9$ and $91.0 S c m^{2} m o l^{- 1}$ respectively. $Λ_{m}^{\circ}$ for $C H_{3} COO H$ will be

$180.5 S c m^{2} m o l^{- 1}$

$290.8 S c m^{2} m o l^{- 1}$

$390.5 S c m^{2} m o l^{- 1}$

$425.5 S c m^{2} m o l^{- 1}$