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If the freezing point of 0.1 molal HA(aq) is –0.2046°C, then pH of the solution is [Kf(H2O) = 1.86°C mol–1 kg] ?
Correct answer is '2'. Can you explain this answer?
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If the freezing point of 0.1 molal HA(aq) is –0.2046°C, then...
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If the freezing point of 0.1 molal HA(aq) is –0.2046°C, then...
Unknown and the cryoscopic constant of the solvent is Kf = 1.86 °C/m, the freezing point depression can be calculated using the equation:
ΔTf = Kf × molality
Assuming that HA dissociates completely in water, the molality can be calculated as follows:
molality = moles of solute / mass of solvent (in kg)
For 0.1 molal HA(aq), the moles of solute in 1 kg of solvent can be calculated as:
moles of solute = 0.1 mol / 1 kg = 0.1 mol/kg
Therefore, the freezing point depression can be calculated as:
ΔTf = 1.86 °C/m × 0.1 mol/kg = 0.186 °C
If the freezing point of the pure solvent is known, the freezing point of the solution can be calculated by subtracting the freezing point depression from the freezing point of the pure solvent. For example, if the freezing point of the pure solvent is -2.0 °C, then the freezing point of the 0.1 molal HA(aq) solution would be:
freezing point of solution = -2.0 °C - 0.186 °C = -2.186 °C
ΔTf = Kf × molality
Assuming that HA dissociates completely in water, the molality can be calculated as follows:
molality = moles of solute / mass of solvent (in kg)
For 0.1 molal HA(aq), the moles of solute in 1 kg of solvent can be calculated as:
moles of solute = 0.1 mol / 1 kg = 0.1 mol/kg
Therefore, the freezing point depression can be calculated as:
ΔTf = 1.86 °C/m × 0.1 mol/kg = 0.186 °C
If the freezing point of the pure solvent is known, the freezing point of the solution can be calculated by subtracting the freezing point depression from the freezing point of the pure solvent. For example, if the freezing point of the pure solvent is -2.0 °C, then the freezing point of the 0.1 molal HA(aq) solution would be:
freezing point of solution = -2.0 °C - 0.186 °C = -2.186 °C
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If the freezing point of 0.1 molal HA(aq) is –0.2046°C, then pH of the solution is [Kf(H2O) = 1.86°C mol–1 kg] ?Correct answer is '2'. Can you explain this answer?
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