Question

Part a in an effusion experiment, it was determined that nitrogen gas, n2, effused at a rate 1.812 times faster than an unknown gas. what is the molar mass of the unknown gas? express your answer to four significant figures and include the appropriate units.

Answer 1

The molar mass of the unknown gas is 92.0 g/mole

Explanation:

Part a in an effusion experiment, it was determined that nitrogen gas, n2, effused at a rate 1.812 times faster than an unknown gas. what is the molar mass of the unknown gas? express your answer to four significant figures and include the appropriate units.

The molar mass is the mass of a sample of that compound divided by the amount of substance in that sample in moles. Molar mass and molecular weight are often confused, but their values are very different.

Graham's law states the rate of diffusion or of effusion of a gas is inversely proportional to  its molecular weight of the square root .

Graham's Law = rate of effusion of gas 2  / rate of effusion of gas 2

Graham's Law = square root (MM1 / MM2), where MM1 and MM2 are the molar masses of the gases.  

Gas_2 = N_2, therefore

1.812 =  √ (MM_1 / MM N_2)

1.812 = √ (MM_1 / 28.0)

1.812 = √ (MM_1 / √ 28.0 )

1.812 = √ (MM_1 / 5.29 )

1.812 x 5.29 = √ MM_1 = 9.59

MM_1 = 9.59^2 = 92.0 g/mole

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Answer 2

Effusion rate, R, is inversely proportional to the square root of the molar mass, M:

R α 1 / √M

Let Ra be the effusion rate of a gas with molar mass Ma and Rb the efussion rate of a gas with molar mass Mb, then:

Ra / Rb = √ [Mb / Ma]

You know Ra / Rb = 1.82 and Ma = molar mass of N2 = 28.02 g/mol

=> Mb = [Ra / Rb]^2 * Ma = (1.82)^2 * 28.02 g/mol = 92.81 g/mol

Answer: 92.81 g/mol