Question #60299

1 Answer
Nov 15, 2015

Answer:

#146"g""/""mol"#

Explanation:

Graham's Law states that the rate of effusion is inversely proportional to the square root of the molar mass:

#Rprop(1)/sqrt(M_r)=k/sqrt(M_r)#

For 2 gases we can divide to get:

#R_1/R_2=sqrt((M_2)/(M_1))#

We can express rate as #V/t# so:

#(V_1)/(t)/(V_2)/(t)=sqrt((M_2)/(M_1))#

Since #t# is the same we can write:

#:.(V_1)/(cancel(t)).(cancel(t))/(V_2)=sqrt((M_2)/(M_1))#

#:.(V_1)/(V_2)=sqrt((M_2)/(M_1))#

I'll refer to the unknown gas as gas 1 and assume the #M_r# of Argon to be 40 #rArr#

#4.83/9.23=sqrt((40)/(M_1))#

#:.((4.83)/(9.23))^(2)=40/M_1#

#:.0.274=40/M_1#

#:.M_1=40/0.274=146#