Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids Solution

STEP 0: Pre-Calculation Summary
Formula Used
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)
MA = (WA*MB*PB°)/(PA°*WB)
This formula uses 6 Variables
Variables Used
Molecular Mass of Liquid A - (Measured in Kilogram) - The Molecular Mass of Liquid A in a mixture of immiscible liquids.
Weight of Liquid A - (Measured in Kilogram) - The Weight of Liquid A in a mixture of immiscible liquids.
Molecular Mass of Liquid B - (Measured in Kilogram) - The Molecular Mass of Liquid B in a mixture of immiscible liquids.
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.
Vapor Pressure of Pure Component A - (Measured in Pascal) - Vapor Pressure of Pure Component A is the pressure exerted by a liquid or solid molecules of only A in a closed system in which they are in equilibrium with the vapour phase.
Weight of Liquid B - (Measured in Kilogram) - The Weight of Liquid B in a mixture of immiscible liquids.
STEP 1: Convert Input(s) to Base Unit
Weight of Liquid A: 0.5 Gram --> 0.0005 Kilogram (Check conversion ​here)
Molecular Mass of Liquid B: 31.8 Gram --> 0.0318 Kilogram (Check conversion ​here)
Vapor Pressure of Pure Component B: 0.25 Pascal --> 0.25 Pascal No Conversion Required
Vapor Pressure of Pure Component A: 2.7 Pascal --> 2.7 Pascal No Conversion Required
Weight of Liquid B: 0.1 Gram --> 0.0001 Kilogram (Check conversion ​here)
STEP 2: Evaluate Formula
Substituting Input Values in Formula
MA = (WA*MB*PB°)/(PA°*WB) --> (0.0005*0.0318*0.25)/(2.7*0.0001)
Evaluating ... ...
MA = 0.0147222222222222
STEP 3: Convert Result to Output's Unit
0.0147222222222222 Kilogram -->14.7222222222222 Gram (Check conversion ​here)
FINAL ANSWER
14.7222222222222 14.72222 Gram <-- Molecular Mass of Liquid A
(Calculation completed in 00.020 seconds)

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Immiscible Liquids Calculators

Partial Vapour Pressure of Immiscible Liquid given Partial Pressure of other Liquid
​ LaTeX ​ Go 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)
Ratio of Partial Vapour Pressures of 2 Immiscible Liquids given Weight and Molecular Mass
​ LaTeX ​ Go 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)
Total Pressure of Mixture of Two Immiscible Liquids
​ LaTeX ​ Go Total Pressure of Mixture of Immiscible Liquids = Vapor Pressure of Pure Component A+Vapor Pressure of Pure Component B
Ratio of Partial Pressure of 2 Immiscible Liquids given Number of Moles
​ LaTeX ​ Go Ratio of Partial Pressures of 2 Immiscible Liquids = Number of Moles of Liquid A/Number of Moles of Liquid B

Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids Formula

​LaTeX ​Go
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)
MA = (WA*MB*PB°)/(PA°*WB)

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 Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids?

Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids calculator uses 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) to calculate the Molecular Mass of Liquid A, The Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids considering A as unknown parameter and B as known parameter. Molecular Mass of Liquid A is denoted by MA symbol.

How to calculate Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids using this online calculator? To use this online calculator for Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids, enter Weight of Liquid A (WA), Molecular Mass of Liquid B (MB), Vapor Pressure of Pure Component B (PB°), Vapor Pressure of Pure Component A (PA°) & Weight of Liquid B (WB) and hit the calculate button. Here is how the Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids calculation can be explained with given input values -> 14.72222 = (0.0005*0.0318*0.25)/(2.7*0.0001).

FAQ

What is Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids?
The Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids considering A as unknown parameter and B as known parameter and is represented as MA = (WA*MB*PB°)/(PA°*WB) or 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). The Weight of Liquid A in a mixture of immiscible liquids, The Molecular Mass of Liquid B in a mixture of immiscible liquids, 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 A is the pressure exerted by a liquid or solid molecules of only A in a closed system in which they are in equilibrium with the vapour phase & The Weight of Liquid B in a mixture of immiscible liquids.
How to calculate Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids?
The Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids considering A as unknown parameter and B as known parameter is calculated using 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). To calculate Molecular Mass of Liquid in Mixture of Two Immiscible Liquids given Weight of Liquids, you need Weight of Liquid A (WA), Molecular Mass of Liquid B (MB), Vapor Pressure of Pure Component B (PB°), Vapor Pressure of Pure Component A (PA°) & Weight of Liquid B (WB). With our tool, you need to enter the respective value for 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 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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