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a conductivity cell when filled with 0.05M solution of KCl records a resistance of 410.5 ohm at 298 K. When filled with CaCl_{2 }soultion (11 g in 500 ml), it records a resistance of 990 ohms. If specific conductance of 0.05M KCl is 0.00189ohm ^{-1 }cm^{ -1 }calculate(i) equivalent conductance of CaCl_{2 }solution

Specific conductance = (1 / Resistance) $\times $ cell constant (KCl solution)

0.00189 = (1/410.5) $\times $ cell constant

Cell constant = 0.7758 cm

^{-1}

Cell constant will be same for both solutions

Specific conductance = (1 / Resistance) $\times $cell constant (CaCl

_{2}solution)

= (1 / 990) $\times $ 0.7758

0.000784 S cm

^{-1}

Equivalent conductance = 1000 $\times $ Specific Conductance / C

where C is the Equivalent concentration or Normality of CaCl

_{2}solution

Normality = Number of gram-equivalents / Volume of solution (in L)

Number of gram-equivalents = mass of CaCl

_{2}/ equivalent-mass

Equivalent Mass = 111 / 2

Equivalent Mass = 55.5 g

Number of gram-equivalents = 5 / 55.5

= 0.09

Normality = 0.09 / 0.5

= 0.18 N

Equivalent conductance = 1000$\times $0.000784 / 0.18

= 4.355 S cm

^{2}eq

^{-1}

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