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jonny [76]
3 years ago
14

80cm^3 of oxygen gas diffussed through a porous holebin 50 seconds how long will it take 120cm^3 of nitrogen (iv) oxide to diffu

se through thr same hole under the same confitions​
Chemistry
1 answer:
iogann1982 [59]3 years ago
6 0

Answer:

time for NO₂ to diffuse = 90 s

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 = \sqrt{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) =  \sqrt{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) =  \sqrt{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) = \sqrt{46 g/mol/32 g/mol}

(1.6 cm³/s)/(120 cm³ / time for NO₂) = \sqrt{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

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