Question

80cm^3 of oxygen gas diffussed through a porous holebin 50 seconds how long will it take 120cm^3 of nitrogen (iv) oxide to diffuse through thr same hole under the same confitions​

Answer 1

time for NO₂ to diffuse = 90 s

Step-by-step explanation:

Data Given:

Amount of O₂ gas transferred = 80 cm³

time of O₂ gas to diffuse = 50s

Amount of NO₂ gas transferred = 120 cm³

time of NO₂ gas to diffuse = ?

Solution:

As we know

rate of diffusion = Amount of gas transferred / time . . . . . . . (1)

we also know that Graham's Law is

rate of diffusion gas A/rate of diffusion gas B =
`√(mB/mA)` . . . (2)

where

mA = molar mass of gas A

mB = molar mass of gas B

combine both equation 1 and 2

(Amount of gas A transferred / time for gas A) / (Amount of gas B transferred / time for gas B) =
`√(mB/mA)` . . . . . . . . (3)

we can write equation 3 for oxygen and nitrogen (iv) oxide

(Amount of gas O₂ transferred / time for gas O₂) / (Amount of gas NO₂ transferred / time for gas NO₂) =
`√(m NO₂/mO₂)` . . . . . . . (4)

• molar mass of O₂ = 2 (16) = 32 g/mol

• molar mass of NO₂ = 14 + 2(16) = 46 g/mol

Put values in equation equation 4

(80 cm³/50 s) / (120 cm³ / time for NO₂) =
`√(46 g/mol/32 g/mol)`

(1.6 cm³/s)/(120 cm³ / time for NO₂) =
`√(1.44)`

(1.6 cm³/s) / (120 cm³ / time for NO₂) = 1.2

Rearrange the above equation

time for NO₂ = 1.2 /(1.6 cm³/s) x 120 cm³

time for NO₂ = 90 s

So,

time for NO₂ to diffuse = 90 s