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प्रश्न
The resistance of a conductivity cell with a 0.1 M KCl solution is 200 ohm. When the same cell is filled with a 0.02 M NaCl solution, the resistance is 1100 ohm. If the conductivity of 0.1 M KCl solution is 0.0129 ohm-1 cm-1, calculate the cell constant and molar conductivity of 0.02 M NaCl solution.
उत्तर
Given: `"R"_((0.1 "M" "KCl"))` = 200 ohm, `"R"_((0.02 "M" "NaCl"))` = 1100 ohm
`"K"_((0.1 "M" "KCl"))` = 0.0129 ohm-1 cm-1
`"G"_((0.02 "M" "NaCl"))` = ?, ∧m = ?
Cell constant (G) = Conductivity (K) × Resistance (R)
= 200 × 0.0129
= 2.58 cm-1
Conductivity `(lambda_(0.02 "M" "NaCl")) = "Cell constant (G)"/("Resistance" ("R"_(0.02 "M" "NaCl")))`
`= (2.58)/1100`
= 2.345 × 10-3 s cm-1
Molar conductivity (λm) `= "Conductivity (K) × 1000"/"Molarity"`
`= (2.345 xx10^-3 xx 1000)/(0.02)`
= 117.25 s m2 mol-1
Hence, the cell constant and moral conductivity of 0.02 M NaCl will be 2.345 × 10-3 s cm-1 and 117.25 s m2 mol-1, respectively.
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