A given volume of methane diffuses in 20s. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions?
[C = 12, H = 1, S = 32, O = 16

Myschool Nuel

1 year ago

Thank you all for your contributions. However, kindly note that there is a relationship between rate and time, so don't mistake or substitute one for the other.

N.B: Rate is inversely proportional to time:
Rate is the change in a component over time, while time is how long an occurrence lasts. As time increases, the rate of a chemical reaction usually decreases. For example, if the rate of a chemical reaction doubles, the time it takes to produce the same amount of product is halved.
Consequently, the formula given in the explanation provided is the correct one but if asked to find rate of diffusion of the gas, you'll use \(r_{1}/r_{2}=\sqrt{M_{2}/M_{1}}\) and if asked to find time or how long, you'll use the formula \(t_{1}/t_{2}=\sqrt{M_{1}/M_{2}}\).

Thanks

Adahomah

2 years ago

The answer is 10

tyjesma

11 years ago

@ equal volumes, t2/t1=√d2/√d1...(Sulphur-iv-oxide)SO2=64, methane CH4=16...t2=?, t1=20s...now subt. D values to get yur answer

Anela

1 year ago

T2/T1 = √m.m2/√m.m1
T2=?
T1=20
m.m2= SO2=32+32=64
m.m1 = CH4=12+4 =16

x/20 = √64/√16
x/20 = 8/4
CROSS MULTIPLY ✖️

4x = 160
DBS BY 4
x = 40 secs.(ans)

BrightTehingbola

2 months ago

i don't understand

Khadeejahhhhhh

1 year ago

Following Graham's law of diffusion;
T1÷T2 = √M2÷√M1 ,
where T represents the time taken to diffuse and M is molar mass. In this question T1 is 20 seconds , M2 is 64,M1 is 16 and T2 is unknown.
Therefore, 20 ÷ T2= √4,
20 ÷ T2 = 2
T2 = 20 ÷ 2
so T2 equals 10 seconds.