A sample of N2O effuses from a container in 43 seconds. How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?

It would take 103s.

Graham’s Law states that :

##”RatepropI/sqrt(M_r)##

Where ##M_r## is the relative molecular mass.

For two gases:

##R_1prop(1)/(sqrtM_(r(1))## ##color(red)((1))##

and:

##R_2prop1/sqrtM_(r(2))## ##color(red)((2))##

Dividing ##color(red)((1))”by”color(red)((2))## gives:

##R_1/R_2=sqrt(M_(r(2))/(M_(r(1))## ##color(red)((3))##

The rate can be written as :

##R=V/t##

I am assuming here that “same amount” means equal volumes:

Since ##V## is the same for both gases we can write:

##R_1=V/t_1##

And:

##R_2=V/t_2##

If we substitute these into ##color(red)((3))## we get:

##t_2/t_1=sqrt(M_(r(2))/(M_(r(1))##

So:

##t_2=sqrt((M_(r(2)))/(M_(r(1))####xxt_1##

##M_(r(1))=M_r[N_2O]=(2xx14)+16=44##

##M_(r(2))=M_r[I_2]=(2xx127)=254##

So:

##t_2=sqrt((254)/(44))xx43=103″s”##

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