Vapor Pressure Calculator for a Solution Containing 24.5 g Solute
Use Raoult law to estimate solution vapor pressure, pressure lowering, and solvent mole fraction.
How to Calculate the Vapor Pressure of a Solution Containing 24.5 g of Solute
If you need to calculate the vapor pressure of a solution containing 24.5 g of solute, the most common starting point is Raoult law. This law links solution vapor pressure to the mole fraction of the solvent. It is one of the core ideas in physical chemistry and process design because it provides a fast way to estimate how a dissolved component changes evaporation behavior. In practical terms, when a nonvolatile solute is dissolved in a solvent, the solvent vapor pressure drops. That drop affects drying rates, boiling conditions, humidity control, and equilibrium design in laboratory and industrial systems.
The important detail is that a mass value by itself, such as 24.5 g, is not enough to compute vapor pressure. You also need the solute molar mass, solvent mass, solvent molar mass, and pure solvent vapor pressure at the temperature of interest. Once those values are known, the math is straightforward and robust for ideal or near ideal behavior.
Core Equation and Why It Works
For a nonvolatile solute in a single solvent, Raoult law is:
Psolution = Xsolvent × P0solvent
where Xsolvent = nsolvent / (nsolvent + nsolute), and moles are computed from mass and molar mass.
This means the vapor pressure of the solution is the pure solvent vapor pressure multiplied by the solvent mole fraction. If solvent molecules are diluted by dissolved solute, fewer solvent molecules are available at the surface, so escaping tendency decreases, and measured vapor pressure is lower.
Worked Example Using 24.5 g Solute
Suppose your solute mass is 24.5 g, the solute is glucose with molar mass 180.16 g/mol, and your solvent is water. Take 100.0 g water with molar mass 18.015 g/mol, and use pure water vapor pressure at 25 C as 23.8 mmHg.
- Compute solute moles: nsolute = 24.5 / 180.16 = 0.1360 mol
- Compute solvent moles: nsolvent = 100.0 / 18.015 = 5.551 mol
- Compute solvent mole fraction: Xsolvent = 5.551 / (5.551 + 0.1360) = 0.9759
- Compute solution vapor pressure: Psolution = 0.9759 × 23.8 = 23.23 mmHg
- Pressure lowering: deltaP = 23.8 – 23.23 = 0.57 mmHg
So a 24.5 g glucose addition in this setup lowers water vapor pressure by about 2.4 percent. This is exactly the kind of result used in colligative property analysis.
Reference Data Table: Vapor Pressure of Pure Water
Temperature is critical because pure solvent vapor pressure changes strongly with temperature. The table below lists widely used reference values for pure water.
| Temperature (C) | Pure Water Vapor Pressure (kPa) | Pure Water Vapor Pressure (mmHg) |
|---|---|---|
| 20 | 2.34 | 17.5 |
| 25 | 3.17 | 23.8 |
| 30 | 4.24 | 31.8 |
| 40 | 7.38 | 55.3 |
| 50 | 12.35 | 92.6 |
Comparison Table: Same 24.5 g Mass, Different Solutes in 100 g Water at 25 C
The same 24.5 g mass can produce very different vapor pressure drops because mole count depends on molar mass and dissociation behavior. The values below assume ideal behavior for neutral molecules and an effective van’t Hoff factor for ionic species.
| Solute | Molar Mass (g/mol) | Approx. Moles from 24.5 g | Estimated Xsolvent | Estimated Psolution (mmHg) |
|---|---|---|---|---|
| Glucose | 180.16 | 0.136 | 0.9759 | 23.23 |
| Urea | 60.06 | 0.408 | 0.9316 | 22.17 |
| Sodium chloride (effective i about 1.9 in moderate solution) | 58.44 | 0.419 mol formula units | 0.8740 (using effective particle moles) | 20.80 |
Step by Step Method You Can Reuse for Any 24.5 g Problem
1) Confirm whether the solute is nonvolatile or volatile
The simple one line form of Raoult law shown earlier is for a nonvolatile solute. If both components are volatile, you need partial pressure terms for each component:
Ptotal = XA P0A + XB P0B
For school and introductory engineering problems that say a solution contains 24.5 g of dissolved solid, the nonvolatile version is usually intended.
2) Use consistent units
- Mass in grams
- Molar mass in grams per mole
- Pressure in one chosen unit, commonly mmHg or kPa
- Temperature matched to the pressure reference value
Unit mismatches are a major source of wrong answers. For example, using water vapor pressure at 30 C while your problem states 25 C can introduce large error.
3) Convert both masses to moles
Colligative property equations depend on moles, not grams. This is why two different compounds at the same 24.5 g can give very different outcomes.
4) Compute mole fraction precisely
Keep at least four significant digits during intermediate steps. Rounding too early can slightly shift final vapor pressure.
5) Multiply by pure solvent vapor pressure
Once mole fraction is known, multiply by pure solvent vapor pressure at the exact temperature. You can then report pressure lowering as:
deltaP = P0 – Psolution
Common Mistakes When Solving 24.5 g Vapor Pressure Problems
- Using molarity instead of mole fraction when Raoult law is requested.
- Ignoring ionic dissociation for salts when the problem expects effective particle count.
- Forgetting that pure solvent vapor pressure changes with temperature.
- Using solute mole fraction by accident instead of solvent mole fraction.
- Entering mass values without checking that the solvent is still the dominant component.
Advanced Notes for Laboratory and Process Contexts
Real systems can deviate from ideal behavior. At higher concentrations, activity coefficients become important and Raoult law may need correction. For electrolytes, the van’t Hoff factor is often concentration dependent, not a fixed textbook constant. In rigorous process simulations, you may use models such as NRTL, Wilson, or UNIQUAC for liquid phase nonideality, and a full equation of state for vapor phase effects under high pressure. Still, for dilute to moderate classroom conditions, the Raoult framework offers excellent first pass estimates.
Another useful interpretation is water activity. For many dilute aqueous solutions, water activity is close to solvent mole fraction. Since equilibrium relative humidity above a solution is linked to water activity, vapor pressure calculations also support humidity control tasks in storage, environmental chambers, and food or pharmaceutical development.
Why 24.5 g Is a Meaningful Input in Design
In many lab protocols, masses such as 24.5 g appear because they fit stock preparation standards, bottle scales, or target molality batches. When this mass is paired with a fixed solvent amount, vapor pressure prediction helps you quickly estimate:
- Expected evaporation rate changes
- Boiling point shifts and distillation behavior
- Headspace composition tendencies
- Safety considerations tied to volatilization and exposure
Even a modest pressure lowering can matter in tightly controlled environments. For example, environmental testing chambers and pharmaceutical processing steps often require predictable vapor behavior.
Authoritative References and Data Sources
For high quality property values and technical background, use government and university references:
- NIST Chemistry WebBook (.gov) for vapor pressure and thermophysical data.
- USGS Water Science School on vapor pressure (.gov) for water related fundamentals.
- Purdue University Chemistry resources (.edu) for instructional chemistry context.
Final Practical Takeaway
To calculate the vapor pressure of a solution containing 24.5 g solute, always convert mass to moles first, compute the solvent mole fraction, and apply Raoult law with the correct pure solvent vapor pressure at your operating temperature. If your solute is nonvolatile, this method is fast and dependable. If the solute is ionic or volatile, include dissociation or multicomponent terms as needed. The calculator above automates these steps and plots how pressure scales with solvent mole fraction so you can validate your intuition and compare scenarios quickly.