A KCl solution of conductivity 0.14 Sm⁻¹ shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω. The conductivity of the HCl solution is ______ *10⁻²S m⁻¹.
A KCl solution of conductivity 0.14 Sm⁻¹ shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω. The conductivity of the HCl solution is ______ *10⁻²S m⁻¹.
The cell constant (G) is given by G = κ * R, where κ is conductivity and R is resistance.
Since the cell constant is constant: (κ * R)KCl = (κ * R)HCl
0.14 Sm? ¹ * 4.19 Ω = κ_HCl * 1.03 Ω
κ_HCl = (0.14 * 4.19) / 1.03 = 0.569 Sm? ¹
This is equivalent to 56.9 * 10? ² Sm? ¹.
The answer, rounded off, is 57.
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CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry Hydrocarbon 2025
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