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Vapour Pressure of water forming an immiscible mixture with a liquid Calculator

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✖The weight of water in an immiscible mixture with another liquid.ⓘ Weight of water in an immiscible mixture [Wwater
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✖Vapor Pressure of Pure Component B is the pressure exerted by a liquid or solid molecules of only B in a closed system in which they are in equilibrium with the vapour phase.ⓘ Vapor Pressure of Pure Component B [PBo]
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✖The Molecular Mass of Liquid B in a mixture of immiscible liquids.ⓘ Molecular Mass of Liquid B [MB]
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✖The Weight of Liquid B in a mixture of immiscible liquids.ⓘ Weight of Liquid B [WB]
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✖The Molecular Mass of Water is equal to 18g per mole.ⓘ Molecular Mass of Water [Mwater]
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✖The Partial Pressure of pure Water is the pressure exerted by water vapours prior addition to solution.ⓘ Vapour Pressure of water forming an immiscible mixture with a liquid [Powater]
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Vapour Pressure of water forming an immiscible mixture with a liquid Solution

STEP 0: Pre-Calculation Summary
Formula Used
Partial Pressure of pure Water = (Weight of water in an immiscible mixture*Vapor Pressure of Pure Component B*Molecular Mass of Liquid B)/(Weight of Liquid B*Molecular Mass of Water)
Powater = (Wwater*PBo*MB)/(WB*Mwater)
This formula uses 6 Variables
Variables Used
Partial Pressure of pure Water - (Measured in Pascal) - The Partial Pressure of pure Water is the pressure exerted by water vapours prior addition to solution.
Weight of water in an immiscible mixture - (Measured in Kilogram) - The weight of water in an immiscible mixture with another liquid.
Vapor Pressure of Pure Component B - (Measured in Pascal) - Vapor Pressure of Pure Component B is the pressure exerted by a liquid or solid molecules of only B in a closed system in which they are in equilibrium with the vapour phase.
Molecular Mass of Liquid B - (Measured in Kilogram) - The Molecular Mass of Liquid B in a mixture of immiscible liquids.
Weight of Liquid B - (Measured in Kilogram) - The Weight of Liquid B in a mixture of immiscible liquids.
Molecular Mass of Water - (Measured in Kilogram) - The Molecular Mass of Water is equal to 18g per mole.
STEP 1: Convert Input(s) to Base Unit
Weight of water in an immiscible mixture: 0.1 Kilogram --> 0.1 Kilogram No Conversion Required
Vapor Pressure of Pure Component B: 10 Pascal --> 10 Pascal No Conversion Required
Molecular Mass of Liquid B: 25 Gram --> 0.025 Kilogram (Check conversion here)
Weight of Liquid B: 0.1 Kilogram --> 0.1 Kilogram No Conversion Required
Molecular Mass of Water: 18 Gram --> 0.018 Kilogram (Check conversion here)
STEP 2: Evaluate Formula
Substituting Input Values in Formula
Powater = (WwaterBo*MB)/(WB*Mwater) --> (0.1*10*0.025)/(0.1*0.018)
Evaluating ... ...
Powater = 13.8888888888889
STEP 3: Convert Result to Output's Unit
13.8888888888889 Pascal -->0.00013707267593278 Standard Atmosphere (Check conversion here)
FINAL ANSWER
0.00013707267593278 Standard Atmosphere <-- Partial Pressure of pure Water
(Calculation completed in 00.047 seconds)
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National Institute of Technology (NIT), Meghalaya
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National Institute of Information Technology (NIIT), Neemrana
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10+ Immiscible Liquids Calculators

Total Vapour Pressure of mixture of given partial pressure of one liquid
Total Pressure of Mixture of Immiscible Liquids = Vapor Pressure of Pure Component B+((Vapor Pressure of Pure Component B*Weight of Liquid A*Molecular Mass of Liquid B)/(Weight of Liquid B*Molecular Mass of Liquid A)) Go
Molecular Mass of a liquid forming an immiscible mixture with Water
Molecular Mass of Liquid B = (Partial Pressure of pure Water*Molecular Mass of Water*Weight of Liquid B)/(Vapor Pressure of Pure Component B*Weight of water in an immiscible mixture) Go
Molecular Mass of liquid in mixture of 2 immiscible liquids given weights of liquids
Molecular Mass of Liquid A = (Weight of Liquid A*Molecular Mass of Liquid B*Vapor Pressure of Pure Component B)/(Vapor Pressure of Pure Component A*Weight of Liquid B) Go
Weight of liquid in mixture of 2 immiscible liquids given weight of other liquid
Weight of Liquid A = (Vapor Pressure of Pure Component A*Molecular Mass of Liquid A*Weight of Liquid B)/(Vapor Pressure of Pure Component B*Molecular Mass of Liquid B) Go
Partial Vapour Pressure of immiscible liquid given partial pressure of other
Vapor Pressure of Pure Component A = (Weight of Liquid A*Molecular Mass of Liquid B*Vapor Pressure of Pure Component B)/(Molecular Mass of Liquid A*Weight of Liquid B) Go
Ratio of Weights of 2 Immiscible Liquids forming Mixture
Ratio of Weights of 2 immiscible liquids = (Vapor Pressure of Pure Component A*Molecular Mass of Liquid A)/(Vapor Pressure of Pure Component B*Molecular Mass of Liquid B) Go
Ratio of Molecular Mass of 2 immiscible liquids
Ratio of Molecular Masses of 2 immiscible liquids = (Vapor Pressure of Pure Component B*Weight of Liquid A)/(Vapor Pressure of Pure Component A*Weight of Liquid B) Go
Ratio of partial vapour pressures of 2 Immiscible Liquids in terms of weight and molecular mass
Ratio of Partial Pressures of 2 Immiscible Liquids = (Weight of Liquid A*Molecular Mass of Liquid B)/(Weight of Liquid B*Molecular Mass of Liquid A) Go
Total Pressure of mixture of two Immiscible Liquids
Total Pressure of Mixture of Immiscible Liquids = Vapor Pressure of Pure Component A+Vapor Pressure of Pure Component B Go
Ratio of partial pressure of 2 immiscible liquids in terms their number of moles
Ratio of Partial Pressures of 2 Immiscible Liquids = Number of moles of Liquid A/No. of moles of Liquid B Go

Vapour Pressure of water forming an immiscible mixture with a liquid Formula

Partial Pressure of pure Water = (Weight of water in an immiscible mixture*Vapor Pressure of Pure Component B*Molecular Mass of Liquid B)/(Weight of Liquid B*Molecular Mass of Water)
Powater = (Wwater*PBo*MB)/(WB*Mwater)

How does both liquids contribute to pressure of a mixture of 2 immiscible liquids?

Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapor pressure you measure will simply be the vapor pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapor. The top one is sealing it in. For the purposes of the rest of this topic, we always assume that the mixture is being stirred or agitated in some way so that the two liquids are broken up into drops. At any one time there will be drops of both liquids on the surface. That means that both of them contribute to the overall vapor pressure of the mixture.

How to Calculate Vapour Pressure of water forming an immiscible mixture with a liquid?

Vapour Pressure of water forming an immiscible mixture with a liquid calculator uses Partial Pressure of pure Water = (Weight of water in an immiscible mixture*Vapor Pressure of Pure Component B*Molecular Mass of Liquid B)/(Weight of Liquid B*Molecular Mass of Water) to calculate the Partial Pressure of pure Water, The Vapour Pressure of water forming an immiscible mixture with a liquid considering A as water and B as other liquids for corresponding parameters. Partial Pressure of pure Water is denoted by Powater symbol.

How to calculate Vapour Pressure of water forming an immiscible mixture with a liquid using this online calculator? To use this online calculator for Vapour Pressure of water forming an immiscible mixture with a liquid, enter Weight of water in an immiscible mixture (Wwater, Vapor Pressure of Pure Component B (PBo), Molecular Mass of Liquid B (MB), Weight of Liquid B (WB) & Molecular Mass of Water (Mwater) and hit the calculate button. Here is how the Vapour Pressure of water forming an immiscible mixture with a liquid calculation can be explained with given input values -> 0.000137 = (0.1*10*0.025)/(0.1*0.018).

FAQ

What is Vapour Pressure of water forming an immiscible mixture with a liquid?
The Vapour Pressure of water forming an immiscible mixture with a liquid considering A as water and B as other liquids for corresponding parameters and is represented as Powater = (Wwater*PBo*MB)/(WB*Mwater) or Partial Pressure of pure Water = (Weight of water in an immiscible mixture*Vapor Pressure of Pure Component B*Molecular Mass of Liquid B)/(Weight of Liquid B*Molecular Mass of Water). The weight of water in an immiscible mixture with another liquid, Vapor Pressure of Pure Component B is the pressure exerted by a liquid or solid molecules of only B in a closed system in which they are in equilibrium with the vapour phase, The Molecular Mass of Liquid B in a mixture of immiscible liquids, The Weight of Liquid B in a mixture of immiscible liquids & The Molecular Mass of Water is equal to 18g per mole.
How to calculate Vapour Pressure of water forming an immiscible mixture with a liquid?
The Vapour Pressure of water forming an immiscible mixture with a liquid considering A as water and B as other liquids for corresponding parameters is calculated using Partial Pressure of pure Water = (Weight of water in an immiscible mixture*Vapor Pressure of Pure Component B*Molecular Mass of Liquid B)/(Weight of Liquid B*Molecular Mass of Water). To calculate Vapour Pressure of water forming an immiscible mixture with a liquid, you need Weight of water in an immiscible mixture (Wwater, Vapor Pressure of Pure Component B (PBo), Molecular Mass of Liquid B (MB), Weight of Liquid B (WB) & Molecular Mass of Water (Mwater). With our tool, you need to enter the respective value for Weight of water in an immiscible mixture, Vapor Pressure of Pure Component B, Molecular Mass of Liquid B, Weight of Liquid B & Molecular Mass of Water and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
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