Answer:
65.14 g
Explanation:
The change in the entropy of two ,metals on mixing is
[tex]\delta S= R[(n_{cu}+n_{Ni})ln(n_{cu}+n_{Ni})-n_{cu}lnn_{cu}-n_{Ni}lnn_{Ni}][/tex]
Here, R is real gas constant, n_{cu} is the number of moles of the copper, and n_{Ni} is the number of moles of nickel
Thus the number of moles of Nickel is
[tex]n_{Ni}=\frac{m_{Ni}}{M_{Ni}}[/tex]
[tex]n_{Ni}=\frac{100}{58.69}[/tex] =1.7 moles
now put n_{Ni} = 1.7  ΔS= 15,
R= 8.314 and solve
[tex]15= R[(n_{cu}+1.7)ln(n_{cu}+1.7)-n_{cu}lnn_{cu}-1.7ln1.7][/tex]
On solving the number of moles of copper in the mixture
n_cu= 1.025
Thus the mass of copper is = n_cuĂ—M_cu = 1.025Ă—63.55 = Â 65.14 g
Therefore the required mass is = Â 65.14 g