If 30cm3 of oxygen diffuses through a porous pot in 7 seconds, how long will...

If 30cm3 of oxygen diffuses through a porous pot in 7 seconds, how long will it take 60cm3 of chlorine to diffuse through the same pot, if the vapour densities of oxygen and chlorine are 16 and 36 respectively?
  • A. 9.3 sec
  • B. 14 sec
  • C. 21 sec
  • D. 28 sec
  • E. 30.3 sec
Correct Answer: Option C
The rate of diffusion of a gas is proportional to the square root of the density.
16 and 7 secs
36 or 6 will be proportional to
7/16 x 6
7/4 x 6
But volume of chlorine is twice that of oxygen. The rate of escape of chlorine is
2/1 x 7/4 x 6/1 = 7 x 3; = 21 sec

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Contributions (23)

4 years ago
30cm3 of O2 diffused in 7sec

60cm3 of O2 will diffused in 60*7 / 30 = 14sec

Given: V.D of O2 (p1) = 16, Cl (p2) = 36

Calculated: t1 = 14sec, t2 = ?

Recall Graham's law of diffusion

\|p2 t2. \|36 t2. 6 t2

----- = --- i.e. ------ = ---- »» --- = ---

\|p1 t1. \|16 14. 4 14

Cross multiplying

t2*4 = 14*6

t2*4 = 84

t2 = 84



t2 = 21sec

Therefore 60cm3 of Chlorine will diffuse in 21seconds

NOTE: \| means Square root

---- means division (all over)

/ also mean division

p (eg p1, p2) means density
1 year ago
v1-30cm3,v2-60cm3 t1-7sec,t2-?,
for rate of gases,R1-V1/T1=30/7=4.2
R2=V2/T2=60/X ,BUTR1/R2=square root of M2/M1
=4.2/60/X=SQUARE ROOT OF 36/16
5 years ago
Still nid berra explanation plsss
Adeshina Ajoye
6 years ago
U All Are wrong
7 years ago
The rate of diffusion of gases is inversely proportional to the square root of it's density
6 years ago
to me..rate of difusion=volume constant/time..there4


30/7 divided by 60/t=square root of (16/36)

t/21=4/6..there4..t=84/6=14..so; t=14secs
Ncay swits
6 years ago
thanks am really been clarified here
3 years ago
The selected answer is wrong:

r1. √m2

--- = ----

r2. √m1

30/7. √16

-----= ----

60/t. √36


REF: new school chemistry
7 years ago
dalton's law of diffusion of gases
7 years ago
Even without calculating it,i am sure of the answer,even in my dream,i would dish out the answer.. . . . . Lol
7 years ago
hy guyz d rate of diffusion is inversly proportnal 2 its vapour density nao.wich dn implies: r1/r2 =(v.d2)^1/2 /(v.d1)^1/2,

r1=7secs, r2=X, v.d1=16, v.d2=36,

7/X = ((36)^1/2)/(16)^1/2

7/X = 6/4

7/X = 3/2

Cros multiply

3X = 14

X = 4.666secs

due 2 d volume ratio of chlorine 2 oxygen, i.e 2:1, d rate of diffusion of 60cm3 of chlorine tru d same porous pot = 4.666secs x 2 = 9.3secs.
  • Festus maro: Rate nd tym r diff..r=1/t
    Like 0    Dislike 0   6 years ago
  • emili: Thanks so much, ur explanation makes much more sense
    Like 0    Dislike 0   6 years ago
  • movicsyn: i Vehemently disagree buh he tried at least...


    Grahams law of diffusion must b obeyed here...the law states dat "gases difuse at d rate which are inversely proportional to d square root of their vapour density"....i.e R=1/ √ V.d....(R = 1 divided by square root of vapour density)..take note that 2 X v.p =relative molecular mass.. And Rate of difusion= 1/time...

    In summary, d working goes thus:.

    30cm3 of Oxygen difuses in 7secs

    60cm3 of Oxygen will difuse in 14 secs..

    Definately, Oxygen Time = 14secs, and its relative molar mass= 32..(2 X its V.d,..2X16)

    FOR chlorine, Chlorine time is unknown= ? And its R.M.M is 72...(2X36)....

    .....applying dis equation..Oxygen time/ chlorine time..= √oxygen relative molecular mass/ √ chlorine R.M.M...

    I.e equate it....(14/?)=(√32/√72)....the unknown ans wil give us 21 secs...

    ........hold pen to understand and feel free to ask if not clear enough!
    Like 1    Dislike 0   5 years ago
  • Chikara44: Nice explanation
    Like 0    Dislike 0   2 years ago
6 years ago
Using grahams law R1 per R2 = square root of P2 per P1...I got 20.7 rounded to 21, not so pleased with my workings tho
7 years ago
Dat is not so helfpul funshofummi

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