A sample of N2O effuses from a container in 47s . Required:
How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?

112.92 s.

Explanation:

Let M₁ be the molar mass of N₂O

Let t₁ be the time taken for N₂O to effuse.

Let M₂ be the molar mass of I₂

Let t₂ be the time taken for I₂ to effuse.

Molar mass (M₁) of N₂O = (14×2) + 16 = 28 + 16 = 44 g/mol

Time (t₁) of effusion of N₂O = 47 s

Molar mass (M₂) of I₂ = 127 × 2 = 254 g/mol

Time (t₂ ) of effusion of I₂ =?

The time take for the same amount of I₂ to effuse can be obtained as follow:

t₂/t₁ = √(M₂/M₁)

t₂/47 = √(254 / 44)

Cross multiply

t₂ = 47 × √(254 / 44)

t₂ = 112.92 s

Therefore, it will take 112.92 s for the same amount of I₂ to effuse.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.