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Lowering of vapour pressure of an aqueous solution of a non-volatile non-electrolyte 1 molal aqueous solution at 100°C is
- a)14.12 torr
- b)312 torr
- c)13.44 torr
- d)352 torr
Correct answer is option 'C'. Can you explain this answer?
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Lowering of vapour pressure of an aqueous solution of a non-volatile n...
(c) Molality = 1 mol kg-1 (solvent) Thus, solute = 1 mole
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Lowering of vapour pressure of an aqueous solution of a non-volatile n...
The lowering of vapor pressure of an aqueous solution of a non-volatile non-electrolyte can be calculated using Raoult's Law.
Raoult's Law states that the vapor pressure of a component in a solution is proportional to its mole fraction in the solution. For a non-volatile non-electrolyte, the vapor pressure of the solvent is not affected by the presence of the solute.
In a 1 molal aqueous solution, the mole fraction of the solute is given by:
X_solute = moles of solute / (moles of solvent + moles of solute)
Since the solute is non-volatile and non-electrolyte, its mole fraction is very small and can be considered negligible compared to the mole fraction of the solvent.
Therefore, the mole fraction of the solvent can be approximated to 1.
According to Raoult's Law, the vapor pressure of the solvent in the solution is equal to the vapor pressure of the pure solvent. At 100°C, the vapor pressure of pure water is 1 atm.
Hence, the lowering of vapor pressure of the aqueous solution is 0 atm.
Raoult's Law states that the vapor pressure of a component in a solution is proportional to its mole fraction in the solution. For a non-volatile non-electrolyte, the vapor pressure of the solvent is not affected by the presence of the solute.
In a 1 molal aqueous solution, the mole fraction of the solute is given by:
X_solute = moles of solute / (moles of solvent + moles of solute)
Since the solute is non-volatile and non-electrolyte, its mole fraction is very small and can be considered negligible compared to the mole fraction of the solvent.
Therefore, the mole fraction of the solvent can be approximated to 1.
According to Raoult's Law, the vapor pressure of the solvent in the solution is equal to the vapor pressure of the pure solvent. At 100°C, the vapor pressure of pure water is 1 atm.
Hence, the lowering of vapor pressure of the aqueous solution is 0 atm.
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Lowering of vapour pressure of an aqueous solution of a non-volatile non-electrolyte 1 molal aqueous solution at 100°C isa)14.12 torrb)312 torrc)13.44 torrd)352 torrCorrect answer is option 'C'. Can you explain this answer?
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