Question

A conductivity cell when filled with 0.05m solution of kcl records a resistance of 410.5 ohm at 25c. When filled with cacl2 solution (11 g in 500 ml), it records a resistance of 990 ohms. If specific conductance of 0.05m kcl is 0.00189 ohm per cm, calculate the equivalent conductance and molar conductance.

Answer 1

  The equivalent conductance and molar conductance is 4.355 S cm2 eq-1

• Specific conductance or conductivity = conductance × cell constant
• Specific conductance = (1 / Resistance) × cell constant      (KCl solution)
• 0.00189 = (1/410.5) × cell constant
• Cell constant = 0.7758 cm-1
• Cell constant will be same for both solutions
• Specific conductance = (1 / Resistance) ×cell constant   (CaCl2 solution)  = (1 / 990) × 0.7758 =0.000784 S cm-1
• Equivalent conductance = 1000 × Specific Conductance / C
• Where C is the Equivalent concentration or Normality of CaCl2 solution
• Normality = Number of gram-equivalents / Volume of solution (in L)
• Number of gram-equivalents = mass of CaCl2 / equivalent-mass
• Equivalent Mass = 111 / 2 =55.5 g
• Number of gram-equivalents = 5 / 55.5  = 0.09
• Normality = 0.09 / 0.5  = 0.18 N
• Equivalent conductance = 1000××0.000784 / 0.18  = 4.355 S cm2 eq-1

Answer 2

Answer:

0.7758

Explanation:

:)

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