At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that

Assessment: JAMB UTME - Chemistry - 2024 Subject: Chemistry

Question 1 Report

At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that

Answer Details

To solve the problem, we can use **Graham's law of effusion**. This law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is represented as:


Rate of diffusion of Gas X / Rate of diffusion of Gas Y = sqrt(Molar mass of Gas Y / Molar mass of Gas X)


According to the given information, gas X diffuses **twice as fast** as gas Y. This implies:


2 = sqrt(Molar mass of Gas Y / Molar mass of Gas X)


To eliminate the square root, square both sides of the equation:


(2)^2 = Molar mass of Gas Y / Molar mass of Gas X


This simplifies to:


4 = Molar mass of Gas Y / Molar mass of Gas X


Rearranging the equation, we find:


Molar mass of Gas Y = 4 * Molar mass of Gas X


This means that **Gas Y is four times as heavy as Gas X**. Therefore, the correct statement is:


  • **Gas Y is four times as heavy as Gas X**

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