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If 200cm of hydrogen diffuses in to second, how will it take for 490on of a gas if x whose vapour density is 25 to diffuse under the Same Condition Rmm hydrogen - 2 ​

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Answer:

Sure, I can help you with that.

The given conditions are:

200 cm of hydrogen diffuses in 2 seconds

Vapour density of gas X = 25

RMM of hydrogen = 2

We need to determine the time it takes for 490 cm³ of gas X to diffuse under the same conditions.

The formula for Graham's Law of Diffusion is:

r1/r2 = √(M2/M1)

where:

r1 is the rate of diffusion of gas 1

r2 is the rate of diffusion of gas 2

M1 is the molar mass of gas 1

M2 is the molar mass of gas 2

In this case, gas 1 is hydrogen and gas 2 is X.

The molar mass of hydrogen is 2 g/mol and the molar mass of X is 25 g/mol.

Plugging these values into the formula, we get:

r1/r2 = √(M2/M1) = √(25/2) = 5/2

We know that 200 cm³ of hydrogen diffuses in 2 seconds. So, the rate of diffusion of hydrogen is 200/2 = 100 cm³/s.

The rate of diffusion of gas X is 5/2 * 100 cm³/s = 250 cm³/s.

Therefore, it will take 490/250 = 1.96 seconds for 490 cm³ of gas X to diffuse under the same conditions.

Hydrogen: Molar mass = 2 g/mol, Rate of diffusion = 100 cm³/s, Time = 2 seconds.

- Gas X: Molar mass = 25 g/mol, Rate of diffusion = 250 cm³/s, Time = 1.96 seconds.

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