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Equivalent, Molar Conductivity and Cell Constant

The molar con...

Question

The molar conductances of $N a C l, H C l$ and $C H_{3} C O O N a$ at infinite dilution are $126.45, 426.16$ and $91 o h m^{- 1} c m^{2} m o l^{- 1}$ respectively. The molar conductance of $C H_{3} C O O H$ at infinite dilution is:

201.28 o h m^{- 1} c m^{2} m o l^{- 1}

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390.71 o h m^{- 1} c m^{2} m o l^{- 1}

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698.28 o h m^{- 1} c m^{2} m o l^{- 1}

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540.48 o h m^{- 1} c m^{2} m o l^{- 1}

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Solution

The correct option is B $390.71 o h m^{- 1} c m^{2} m o l^{- 1}$
$Λ_{N a C l} = Λ_{{N a}^{+}} + Λ_{C l^{-}} Λ_{H C l} = Λ_{H^{+}} + Λ_{C l^{-}} Λ_{C H_{3} C O O N a} = Λ_{{N a}^{+}} + Λ_{C H_{3} C O O^{-}} L e t, Λ_{{N a}^{+}} = x, Λ_{{C l}^{-}} = y, Λ_{H^{+}} = z, Λ_{C H_{3} C O O^{-}} = w G i v e n, x + y = 126.45 y + z = 426.16 x + w = 91$
From the above 3 equations, value of $z + w = 390.71$

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Similar questions

The molar conductances of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91 ohm^-1 cm² mol^-1 respectively. The molar conductance of CH₃COOH at infinite dilution is

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o h m^{- 1} o h m^{- 1} c m^{2} m o l^{- 1}

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120 o h m^{- 1} c m^{2} m o l^{- 1}

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198 o h m^{- 1} c m^{2} m o l^{- 1}

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The molar conductances of NaCl,HCl and CH3COONa at infinite dilution are 126.45,426.16 and 91 ohm−1cm2mol−1 respectively. The molar conductance of CH3COOH at infinite dilution is:

The correct option is B 390.71 ohm−1cm2mol−1ΛNaCl=ΛNa++ΛCl−ΛHCl=ΛH++ΛCl−ΛCH3COONa=ΛNa++ΛCH3COO−Let,ΛNa+=x,ΛCl−=y,ΛH+=z,ΛCH3COO−=wGiven,x+y=126.45y+z=426.16x+w=91From the above 3 equations, value of z+w=390.71

The molar conductances of $N a C l, H C l$ and $C H_{3} C O O N a$ at infinite dilution are $126.45, 426.16$ and $91 o h m^{- 1} c m^{2} m o l^{- 1}$ respectively. The molar conductance of $C H_{3} C O O H$ at infinite dilution is: