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Weight of Water required to form Immiscible Mixture with Liquid given Weight Calculator

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✖The Weight of Liquid B in a mixture of immiscible liquids.ⓘ Weight of Liquid B [W_B]			+10% -10%
✖The Partial Pressure of Pure Water is the pressure exerted by water vapours prior addition to solution.ⓘ Partial Pressure of Pure Water [P^owater]			+10% -10%
✖The Molecular Mass of Water is equal to 18g per mole.ⓘ Molecular Mass of Water [M_water]			+10% -10%
✖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 Molecular Mass of Liquid B in a mixture of immiscible liquids.ⓘ Molecular Mass of Liquid B [M_B]			+10% -10%

✖The Weight of Water in Immiscible Mixture with another liquid.ⓘ Weight of Water required to form Immiscible Mixture with Liquid given Weight [W_water]

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Weight of Water required to form Immiscible Mixture with Liquid given Weight Solution

STEP 0: Pre-Calculation Summary

Formula Used

Weight of Water in Immiscible Mixture = (Weight of Liquid B*Partial Pressure of Pure Water*Molecular Mass of Water)/(Vapor Pressure of Pure Component B*Molecular Mass of Liquid B)
W_water = (W_B*P^owater*M_water)/(P_B°*M_B)
This formula uses 6 Variables

Variables Used

Weight of Water in Immiscible Mixture - (Measured in Kilogram) - The Weight of Water in Immiscible Mixture with another liquid.
Weight of Liquid B - (Measured in Kilogram) - The Weight of Liquid B in a mixture of immiscible liquids.
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.
Molecular Mass of Water - (Measured in Kilogram) - The Molecular Mass of Water is equal to 18g per mole.
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.

STEP 1: Convert Input(s) to Base Unit

Weight of Liquid B: 0.1 Gram --> 0.0001 Kilogram (Check conversion here)
Partial Pressure of Pure Water: 0.53 Pascal --> 0.53 Pascal No Conversion Required
Molecular Mass of Water: 18 Gram --> 0.018 Kilogram (Check conversion here)
Vapor Pressure of Pure Component B: 0.25 Pascal --> 0.25 Pascal No Conversion Required
Molecular Mass of Liquid B: 31.8 Gram --> 0.0318 Kilogram (Check conversion here)

STEP 2: Evaluate Formula

Substituting Input Values in Formula

W_water = (W_B*P^owater*M_water)/(P_B°*M_B) --> (0.0001*0.53*0.018)/(0.25*0.0318)

Evaluating ... ...

W_water = 0.00012

STEP 3: Convert Result to Output's Unit

0.00012 Kilogram -->0.12 Gram (Check conversion here)

FINAL ANSWER

0.12 Gram <-- Weight of Water in Immiscible Mixture

(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

Weight of Water required to form Immiscible Mixture with Liquid given Weight Formula

LaTeX Go

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 Weight of Water required to form Immiscible Mixture with Liquid given Weight?

Weight of Water required to form Immiscible Mixture with Liquid given Weight calculator uses Weight of Water in Immiscible Mixture = (Weight of Liquid B*Partial Pressure of Pure Water*Molecular Mass of Water)/(Vapor Pressure of Pure Component B*Molecular Mass of Liquid B) to calculate the Weight of Water in Immiscible Mixture, The Weight of Water required to form Immiscible Mixture with Liquid given Weight considering A as water and B as other liquids for corresponding parameters. Weight of Water in Immiscible Mixture is denoted by W_water symbol.

How to calculate Weight of Water required to form Immiscible Mixture with Liquid given Weight using this online calculator? To use this online calculator for Weight of Water required to form Immiscible Mixture with Liquid given Weight, enter Weight of Liquid B (W_B), Partial Pressure of Pure Water (P^owater), Molecular Mass of Water (M_water), Vapor Pressure of Pure Component B (P_B°) & Molecular Mass of Liquid B (M_B) and hit the calculate button. Here is how the Weight of Water required to form Immiscible Mixture with Liquid given Weight calculation can be explained with given input values -> 152.64 = (0.0001*0.53*0.018)/(0.25*0.0318).

FAQ

What is Weight of Water required to form Immiscible Mixture with Liquid given Weight?

The Weight of Water required to form Immiscible Mixture with Liquid given Weight considering A as water and B as other liquids for corresponding parameters and is represented as W_water = (W_B*P^owater*M_water)/(P_B°*M_B) or Weight of Water in Immiscible Mixture = (Weight of Liquid B*Partial Pressure of Pure Water*Molecular Mass of Water)/(Vapor Pressure of Pure Component B*Molecular Mass of Liquid B). The Weight of Liquid B in a mixture of immiscible liquids, The Partial Pressure of Pure Water is the pressure exerted by water vapours prior addition to solution, The Molecular Mass of Water is equal to 18g per mole, 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.

How to calculate Weight of Water required to form Immiscible Mixture with Liquid given Weight?

The Weight of Water required to form Immiscible Mixture with Liquid given Weight considering A as water and B as other liquids for corresponding parameters is calculated using Weight of Water in Immiscible Mixture = (Weight of Liquid B*Partial Pressure of Pure Water*Molecular Mass of Water)/(Vapor Pressure of Pure Component B*Molecular Mass of Liquid B). To calculate Weight of Water required to form Immiscible Mixture with Liquid given Weight, you need Weight of Liquid B (W_B), Partial Pressure of Pure Water (P^owater), Molecular Mass of Water (M_water), Vapor Pressure of Pure Component B (P_B°) & Molecular Mass of Liquid B (M_B). With our tool, you need to enter the respective value for Weight of Liquid B, Partial Pressure of Pure Water, Molecular Mass of Water, Vapor Pressure of Pure Component B & Molecular Mass 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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