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Ratio of Molecular Mass of 2 Immiscible Liquids Calculator

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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 [P_B°]			+10% -10%
✖The Weight of Liquid A in a mixture of immiscible liquids.ⓘ Weight of Liquid A [W_A]			+10% -10%
✖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.ⓘ Vapor Pressure of Pure Component A [P_A°]			+10% -10%
✖The Weight of Liquid B in a mixture of immiscible liquids.ⓘ Weight of Liquid B [W_B]			+10% -10%

✖The Ratio of Molecular Masses of 2 immiscible liquids in a mixture say liquid A and liquid B.ⓘ Ratio of Molecular Mass of 2 Immiscible Liquids [M_A:B]

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Ratio of Molecular Mass of 2 Immiscible Liquids Solution

STEP 0: Pre-Calculation Summary

Formula Used

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)
M_A:B = (P_B°*W_A)/(P_A°*W_B)
This formula uses 5 Variables

Variables Used

Ratio of Molecular Masses of 2 Immiscible Liquids - The Ratio of Molecular Masses of 2 immiscible liquids in a mixture say liquid A and liquid B.
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.
Weight of Liquid A - (Measured in Kilogram) - The Weight of Liquid A in a mixture of immiscible liquids.
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

Vapor Pressure of Pure Component B: 0.25 Pascal --> 0.25 Pascal No Conversion Required
Weight of Liquid A: 0.5 Gram --> 0.0005 Kilogram (Check conversion here)
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

M_A:B = (P_B°*W_A)/(P_A°*W_B) --> (0.25*0.0005)/(2.7*0.0001)

Evaluating ... ...

M_A:B = 0.462962962962963

STEP 3: Convert Result to Output's Unit

0.462962962962963 --> No Conversion Required

FINAL ANSWER

0.462962962962963 ≈ 0.462963 <-- Ratio of Molecular Masses of 2 Immiscible Liquids

(Calculation completed in 00.004 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

Ratio of Molecular Mass of 2 Immiscible Liquids Formula

LaTeX Go

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)
M_A:B = (P_B°*W_A)/(P_A°*W_B)

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 Ratio of Molecular Mass of 2 Immiscible Liquids?

Ratio of Molecular Mass of 2 Immiscible Liquids calculator uses 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) to calculate the Ratio of Molecular Masses of 2 Immiscible Liquids, The Ratio of Molecular Mass of 2 Immiscible Liquids formula is defined as the ratio of the molecular masses in mixture. Ratio of Molecular Masses of 2 Immiscible Liquids is denoted by M_A:B symbol.

How to calculate Ratio of Molecular Mass of 2 Immiscible Liquids using this online calculator? To use this online calculator for Ratio of Molecular Mass of 2 Immiscible Liquids, enter Vapor Pressure of Pure Component B (P_B°), Weight of Liquid A (W_A), Vapor Pressure of Pure Component A (P_A°) & Weight of Liquid B (W_B) and hit the calculate button. Here is how the Ratio of Molecular Mass of 2 Immiscible Liquids calculation can be explained with given input values -> 0.462963 = (0.25*0.0005)/(2.7*0.0001).

FAQ

What is Ratio of Molecular Mass of 2 Immiscible Liquids?

The Ratio of Molecular Mass of 2 Immiscible Liquids formula is defined as the ratio of the molecular masses in mixture and is represented as M_A:B = (P_B°*W_A)/(P_A°*W_B) or 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). 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 Weight of Liquid A in a mixture of immiscible liquids, 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 Ratio of Molecular Mass of 2 Immiscible Liquids?

The Ratio of Molecular Mass of 2 Immiscible Liquids formula is defined as the ratio of the molecular masses in mixture is calculated using 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). To calculate Ratio of Molecular Mass of 2 Immiscible Liquids, you need Vapor Pressure of Pure Component B (P_B°), Weight of Liquid A (W_A), Vapor Pressure of Pure Component A (P_A°) & Weight of Liquid B (W_B). With our tool, you need to enter the respective value for Vapor Pressure of Pure Component B, Weight of Liquid A, 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.

How many ways are there to calculate Ratio of Molecular Masses of 2 Immiscible Liquids?

In this formula, Ratio of Molecular Masses of 2 Immiscible Liquids uses Vapor Pressure of Pure Component B, Weight of Liquid A, Vapor Pressure of Pure Component A & Weight of Liquid B. We can use 1 other way(s) to calculate the same, which is/are as follows -

Ratio of Molecular Masses of 2 Immiscible Liquids = (Weight of Water in Immiscible Mixture*Vapor Pressure of Pure Component B)/(Partial Pressure of Pure Water*Weight of Liquid B)

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