Answer:
[tex]m_{Ag}=2,265.9g[/tex]
Explanation:
Hello!
In this case, since the definition of entropy in a random mixture is:
[tex]\Delta S=-n_TR\Sigma[x_i*ln(x_i)][/tex]
For this silver-gold mixture we write:
[tex]\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )][/tex]
By knowing the moles of gold:
[tex]n_{Au}=100g*\frac{1mol}{197g} =0.508mol[/tex]
It is possible to write the aforementioned formula in terms of the variable [tex]x[/tex] representing the moles of silver:
[tex]20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )][/tex]
Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:
[tex]x=n_{Ag}=21.0molAg[/tex]
So the mass is:
[tex]m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g[/tex]
Best regards!
