The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 x 103 ohm. Calculate its resistivity, conductivity and molar conductivity.

                                        l = 50 cm                                        A = πr2 = 3.14 × (0.5)2 = 0.785 cm                Cell constant = lA = 500.785= 63.694 cm-1
(i)    Resistivity  = RCell constant = 5.55×10363.694 = 87.135 Ω cm.
(ii)    Conductivity = 1resistivity=187.135 = 1.148 × 10-2s cm-1
(iii)    Molar conductivity m = 103KM = 103×1.148×10-20.05                                m = 229.6 s cm2 mol-1
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Λ°m for NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 s cm2 mol–1respectively. Calculate Λ° for HAC.

Λ° (HAC) = λ°(H+) + λ°(AC)
= λ° (H+) + λ°(Cl) + λ° (AC) + λ° (Na+) – λ°(Cl) – λ°(Na+)
= Λ° (HCl) + Λ° (NaAc) – Λ° (NaCl)
= (425.9 + 91.0 – 126.4) s cm2 mol–1 = 390.5 s cm2 mol–1

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The conductivity of 0.00241 M acetic acid is 7.896 x 10–5 s cm–1. Calculate its molar conductivity and if Λ°m for acetic acid is 390.5 s cm2 mol–1, what is its dissociation constant?


                                      M = 0.00241 M
                                      K = 7.896 x 10-5 s cm-1
                                     m = 103KM = 103×7.896×10-50.00241m = 32.76 s cm2 mol-1
                                       α = m°m = 32.76390.5 = 0.0839K = 21-α = 0.00241×(0.0839)21-0.0839K = 1.85 × 10-5

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Resistance of a conductivity cell filled with 0.1 M KCl solution is 100 £2. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 Ω. Calculate the conductivity and molar conductivity of 0.02 M KC1 solution. The conductivity of 0.1 M KCl at 298 K is 0.0129 s cm–1.

Cell constant = K x R
                     = 0.0129 ohm-1 cm-1×100 ohm
                     = 1.29 cm-1

K(0.2 M KCl) = Cell constantResistance = 1.29520                        = 2.48 × 10-3 s cm-1
m = 103KM = 103×2.48×10-30.02m = 124 s cm2 mol-1

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250 cm3 solution containing 1 g of Ca(NO3)2 in conductivity cell offered a resistance of 48 ohms. The cell constant of conductivity of the solution is 0.56 cm-1. Find the molar conductivity.

                   Can(NO3)2 = 1 g164 g mol-1 = 0.0061 mol                               V = 250 cm3 = 0.250 L

                        M = nV = 0.00610.250=0.0244 MK = Cell constantResistance = 0.5648 = 1.167 × 10-2 ohm-1 cm-1   
   
                         m = 103KM = 103×1.167×10-20.0244 m = 478.28 s cm2 mol-1.
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