### Graham’s Law Calculator:

Enter the Graham’s law Effusion in Rate of effusion of Gas 1 in mol/s, Rate of effusion of Gas 2 in mol/s, Molar mass of Gas 1 in g/mol, Molar mass of Gas 2 in g/mol , Graham’s law Diffusion in Rate of diffusion of Gas 1 in mol/s, Rate of diffusion of Gas 2 in mol/s, Molar mass of Gas 1 in g/mol and you get the Molar mass of Gas 2 in g/mol.

### Graham’s Law Formula:

Graham’s law formula is rate of effusion of the first gas Rate1 and divided by the rate of effusion of the second gas Rate 2 is equal to the root of the molar mass of gas 2 M_{2} (g mol^{-1 ) } in gram mol^{-1} and divided by the molar mass of gas 1 M_{1}(g mol^{-1 )} in gram per mol-1.Hence the Graham’s law formula can be written as

Rate1 / Rate 2 =√ M_{2} (g mol^{-1 )} / M_{1}(g mol^{-1 )}

M_{1} = molar mass of gas 1

M_{2} = molar mass of gas 2

Rate1 = rate of effusion of the first gas

Rate2 = rate of effusion of the second gas

½ m_{1}*v_{1}^{2} =1/2 *m_{2}*V_{2}^{2}

v_{1}^{2} / V_{2}^{2 =}√ M_{2 }/ M_{1}

Rate1 / Rate 2 =√ M_{2 }/ M_{1}

### Example:1

Calculate the Inert gases Ne atoms how fast travel than Xe?

### Answer:

Rate1 / Rate 2 =√ M_{2 }/ M_{1}

=√f131.29 /20.18 =2.55

N_{e} is 2.55 times faster

### Example:2

Calculate the molar mass of the unknown gas? In a unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperatures and pressure.

### Answer:

Rate1 / Rate 2 =√ M_{2 }/ M_{1}

Rate_{unknown } / Rate N_{2} =√ M N_{2} / M_{unknown}

0.5 =√ 28 g mol^{-1} / M_{unknown}

Squaring on both sides

(0.5)^{2} =(√ 28 g mol^{-1} / M_{unknown})^{2}

M_{unknown } =28 /0.25 =112 g mol^{-1}