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Basile [38]
3 years ago
10

A tank containing both HF and HBr gases developed a leak. The ratio of the rate of effusion of HF to the rate of effusion of HBr

is ________.
A. 0.247
B. 2.01
C. 4.04
D. 16.3
E. 0.497
Chemistry
2 answers:
Alchen [17]3 years ago
6 0

Answer: Option (B) is the correct answer.

Explanation:

According to the Graham's law, rate of effusion of a gas is inversely proportional to the square root of molar mass of the gas.

Mathematically,   Rate of effusion = \frac{1}{\sqrt{M}}

where,   M = molar mass of gas

As molar mass of HF is 20.01 g/mol and molar mass of HBr is 80.91 g/mol. Therefore, calculate the ratio of rate of effusion of HF to the rate of effusion of HBr as follows.

        \frac{R_{HF}}{R_{HBr}} = \sqrt{\frac{80.91 g/mol}{20.01 g/mol}}

                    = \sqrt{4.04}  

                    = 2.009

                    = 2.01 (approx)

Thus, we can conclude that ratio of the rate of effusion of HF to the rate of effusion of HBr is 2.01.

exis [7]3 years ago
4 0

Answer:

E. 0.497

Explanation:

Graham's law of diffusion or effusion states that the rate of diffusion or effusion is inversely proportional to the square root of the molecular mass.

k=r√m

1=HF, 2=HBr

Molar mass of HF=20. Molar mass of HBr=80.91

r1√20=r2√80.91

r1/r2=4.472/8.994

r1/r2=0.497

r1√m1=r2√m2

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Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

  • To solve this problem, we can use the mathematical relation:

ν = \sqrt{3RT/M}

Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,

R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,

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<em>∴ The molar mass of the gas is 36.25 g/mol.</em>


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