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Calculate the mass of a non volatile solute ( molecular mass = 40) ,which. should be dissolved in 114g octane to reduce it's vapour pressure to 80%?
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Introduction
To find the mass of a non-volatile solute needed to decrease the vapor pressure of octane, we will apply Raoult's Law.
Given Data
- Molecular mass of solute = 40 g/mol
- Mass of octane (solvent) = 114 g
- Vapor pressure reduction = 80%
Calculating Vapor Pressure
- The initial vapor pressure of pure octane = P0
- Reduced vapor pressure = 80% of P0 = 0.8 P0
Applying Raoult's Law
According to Raoult's Law:
- P_solution = X_solvent * P0
Where:
- P_solution is the vapor pressure of the solution
- X_solvent is the mole fraction of the solvent
Since we know the reduced vapor pressure:
- 0.8 P0 = X_solvent * P0
- Therefore, X_solvent = 0.8
Finding Mole Fraction of Solute
- Mole fraction of solute (X_solute) = 1 - X_solvent
- X_solute = 1 - 0.8 = 0.2
Calculating Moles of Octane
- Molar mass of octane (C8H18) = 114 g/mol
- Moles of octane = 114 g / 114 g/mol = 1 mol
Finding Moles of Solute
Using the mole fraction:
- X_solute = moles of solute / (moles of solute + moles of solvent)
- 0.2 = n / (n + 1)
- Solving gives n (moles of solute) = 0.2 mol
Calculating Mass of Solute
- Mass of solute = moles * molar mass = 0.2 mol * 40 g/mol = 8 g
Conclusion
To reduce the vapor pressure of 114 g of octane to 80%, 8 g of the non-volatile solute should be dissolved.
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Direction: Read the passage given below and answer the following questions: Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapour pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapour pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. Q.When a non volatile solid is added to pure water it will

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Calculate the mass of a non volatile solute ( molecular mass = 40) ,which. should be dissolved in 114g octane to reduce it's vapour pressure to 80%?
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Calculate the mass of a non volatile solute ( molecular mass = 40) ,which. should be dissolved in 114g octane to reduce it's vapour pressure to 80%? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Calculate the mass of a non volatile solute ( molecular mass = 40) ,which. should be dissolved in 114g octane to reduce it's vapour pressure to 80%? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the mass of a non volatile solute ( molecular mass = 40) ,which. should be dissolved in 114g octane to reduce it's vapour pressure to 80%?.
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