Well, the molar mass of palladium is #106.4*g*mol^-1#, and the molar mass of gold is #197.0*g*mol^-1#.
We know that #x_1xx106.4*g*mol^-1+x_2xx197.0*g*mol^-1# #-=67.2*g#. And here #x_1="moles of palladium"#, and #x_2="moles of golds"#.
We further know that #(x_1+x_2)xx6.02214xx10^23=2.49xx10^23#, by the problem's specification. And thus we gots the REQUIRED two equation to solve for TWO UNKNOWNS......
And so #(x_1+x_2)=(2.49xx10^23)/(6.02214xx10^23)=0.414#...
And thus, #x_1=0.414-x_2#, and we substitute this back into #x_1xx106.4*g*mol^-1+x_2xx197.0*g*mol^-1# #-=67.2*g#.
So....#(0.414-x_2)xx106.4*g*mol^-1+x_2xx197.0*g*mol^-1# #-=67.2*g#, and to simplify.....
....#(0.414-x_2)xx106.4+x_2xx197.0=67.2#.
#44.0-106.4*x_2+197.0*x_2=67.2#
#44.0+90.8*x_2=67.2#
#x_2=(67.2-44.0)/(90.8)=0.256#
And back-substituting, #x_1=0.414-0.256=0.158#.
I think that is enuff arifmetick for anyone, but we should check on the consistency of our results......
If we gots #0.256*mol# with respect to gold we got a mass of #0.256*molxx197*g*mol^-1=50.4*g#, and a mass of #0.158*molxx106.42*g*mol^-1=16.81*g# with respect to palladium. The sum of the masses is #67.2*g# as required by the boundary conditions of the problem. Are you happy with this treatment?
