Question

O2(g) effuses at a rate that is ______ times that of br2(g) under the same conditions.

Answer 1

Oxygen gas effuses at a rate that is 2.236 times that of bromine gas under the same conditions.

Step-by-step explanation:

Graham's Law:

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

`\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}`

We are given:

Molar mass of oxygen gas= 32 g/mol

Molar mass of bromine gas = 160 g/mol

By taking their ratio, we get:

`(r_(O_2))/(r_(Br_2))=\sqrt{(160 g/mol)/(32 g/mol)}`

`(r_(O_2))/(r_(Br_2))=√(5)=2.236`

`r_(O_2)=2.236* r_(Br_2)`

Oxygen gas effuses at a rate that is 2.236 times that of bromine gas under the same conditions.

Answer 2

O2 is effuses at a rate that is 3.2 times faster than Br2. This is because the quotient of the molar mass of br2 and molar mass of O2 is ten. The square root of 10 is 3.2