Molar Mass of Second Gas by Graham's law Solution

STEP 0: Pre-Calculation Summary
Formula Used
Molar Mass of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Molar Mass of First Gas
M2 = ((r1/r2)^2)*M1
This formula uses 4 Variables
Variables Used
Molar Mass of Second Gas - (Measured in Kilogram Per Mole) - The Molar Mass of Second Gas is defined as the mass of the gas per mole.
Rate of Effusion of First Gas - (Measured in Cubic Meter per Second) - The Rate of Effusion of First Gas is the special case of diffusion when the first gas is allowed to escape through the small hole.
Rate of Effusion of Second Gas - (Measured in Cubic Meter per Second) - The Rate of Effusion of Second Gas is the special case of diffusion when the second gas is allowed to escape through the small hole.
Molar Mass of First Gas - (Measured in Kilogram Per Mole) - The Molar Mass of First Gas is defined as the mass of the gas per mole.
STEP 1: Convert Input(s) to Base Unit
Rate of Effusion of First Gas: 2.12 Cubic Meter per Second --> 2.12 Cubic Meter per Second No Conversion Required
Rate of Effusion of Second Gas: 0.12 Cubic Meter per Second --> 0.12 Cubic Meter per Second No Conversion Required
Molar Mass of First Gas: 34.56 Gram Per Mole --> 0.03456 Kilogram Per Mole (Check conversion ​here)
STEP 2: Evaluate Formula
Substituting Input Values in Formula
M2 = ((r1/r2)^2)*M1 --> ((2.12/0.12)^2)*0.03456
Evaluating ... ...
M2 = 10.78656
STEP 3: Convert Result to Output's Unit
10.78656 Kilogram Per Mole -->10786.56 Gram Per Mole (Check conversion ​here)
FINAL ANSWER
10786.56 Gram Per Mole <-- Molar Mass of Second Gas
(Calculation completed in 00.007 seconds)

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Graham's Law Calculators

Rate of Effusion for Second Gas by Graham's law
​ LaTeX ​ Go Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Molar Mass of Second Gas/Molar Mass of First Gas))
Rate of Effusion for First Gas by Graham's law
​ LaTeX ​ Go Rate of Effusion of First Gas = (sqrt(Molar Mass of Second Gas/Molar Mass of First Gas))*Rate of Effusion of Second Gas
Molar Mass of Second Gas by Graham's law
​ LaTeX ​ Go Molar Mass of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Molar Mass of First Gas
Molar Mass of First Gas by Graham's law
​ LaTeX ​ Go Molar Mass of First Gas = Molar Mass of Second Gas/((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)

Molar Mass of Second Gas by Graham's law Formula

​LaTeX ​Go
Molar Mass of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Molar Mass of First Gas
M2 = ((r1/r2)^2)*M1

What is Graham's law?

Graham's law of effusion (also called Graham's law of diffusion) was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the molar mass of its particles. Graham's law is most accurate for molecular effusion which involves the movement of one gas at a time through a hole. It is only approximate for diffusion of one gas in another or in air, as these processes involve the movement of more than one gas. In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore, the rates of diffusion of different gases are inversely proportional to the square roots of their mass densities.

How to Calculate Molar Mass of Second Gas by Graham's law?

Molar Mass of Second Gas by Graham's law calculator uses Molar Mass of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Molar Mass of First Gas to calculate the Molar Mass of Second Gas, The Molar mass of second gas by Graham's law formula is defined as rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Molar Mass of Second Gas is denoted by M2 symbol.

How to calculate Molar Mass of Second Gas by Graham's law using this online calculator? To use this online calculator for Molar Mass of Second Gas by Graham's law, enter Rate of Effusion of First Gas (r1), Rate of Effusion of Second Gas (r2) & Molar Mass of First Gas (M1) and hit the calculate button. Here is how the Molar Mass of Second Gas by Graham's law calculation can be explained with given input values -> 1.1E+7 = ((2.12/0.12)^2)*0.03456.

FAQ

What is Molar Mass of Second Gas by Graham's law?
The Molar mass of second gas by Graham's law formula is defined as rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight and is represented as M2 = ((r1/r2)^2)*M1 or Molar Mass of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Molar Mass of First Gas. The Rate of Effusion of First Gas is the special case of diffusion when the first gas is allowed to escape through the small hole, The Rate of Effusion of Second Gas is the special case of diffusion when the second gas is allowed to escape through the small hole & The Molar Mass of First Gas is defined as the mass of the gas per mole.
How to calculate Molar Mass of Second Gas by Graham's law?
The Molar mass of second gas by Graham's law formula is defined as rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight is calculated using Molar Mass of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Molar Mass of First Gas. To calculate Molar Mass of Second Gas by Graham's law, you need Rate of Effusion of First Gas (r1), Rate of Effusion of Second Gas (r2) & Molar Mass of First Gas (M1). With our tool, you need to enter the respective value for Rate of Effusion of First Gas, Rate of Effusion of Second Gas & Molar Mass of First Gas 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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