CHEMISTRY

JAMB 2002 - Question 48

Chemistry 2002 JAMB Past Questions - Question 48: A gas X diffuses twice as fast as gas Y under the same conditions. If the relative molecular mass of X is 28, calculate the relative molecular mass of Y.

Choose the correct answers from the options given.
A gas X diffuses twice as fast as gas Y under the same conditions. If the relative molecular mass of X is 28, calculate the relative molecular mass of Y.
A:
B:
C:
D:
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Correct Answer

A

Explanation

If R is the rate diffusiontherefore X diffuses at 2R andY diffuses at R2R/R = Y/x = Y/28

To find the relative molecular mass of gas Y, we can use Graham's law of diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

The ratio of the rates of diffusion of two gases is given by:

Rate of diffusion of X / Rate of diffusion of Y = √(Molar mass of Y) / √(Molar mass of X)

Given that gas X diffuses twice as fast as gas Y, we can express this as:

2/1 = √(Molar mass of Y) / √(28)

Squaring both sides of the equation gives us:

4 = Molar mass of Y / 28

Multiplying both sides by 28 gives us:

Molar mass of Y = 4 * 28
Molar mass of Y = 112

Therefore, the relative molecular mass of gas Y is 112.
therefore Y/28 =4Y = 112

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