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Water vapor will diffuse faster than butane C4H10. Calcualte in terms of percent how much faster.?
Water vapot has a weight of 18. Butane has a weight of 58.
Intially I set the problem up: rate A water vapor/ rate B butane = sqrt MWB Butane/ MW A water vapor. = 1.80 % faster.
then I did the problem by 18/ 58 sqrt= 55 % faster. This looks more right, but I don't understand how this could be the right anwser if water vpaor is rate A then it should be one the bottom when taking the sqaure root.
1 Answer
- ?Lv 77 years agoFavorite Answer
IMO you are doing it correctly but getting mixed up in the algebra.
The calculation uses Graham's Law of Effusion:
Rate(A)/Rate(B) = √(M(B)/M(A))
http://en.wikipedia.org/wiki/Graham%27s_law
Rate(H2O)/Rate(Butane) = √(M(Butane)/M(H2O))
Rate(H2O)/Rate(Butane) = √(58/18) = √((3.22) = 1.795
Now multiply this out:
Rate(H2O)/Rate(Butane) = 1.795
Rate(H2O) = 1.795*(Rate(Butane))
Look at this expression. We interpret this as meaning that the rate of effusion of H2O is 79.5% greater than the rate of effusion of butane.