Melts in the system pb-sn exhibit regular solution behavior. at 473°c apb = 0.055 in a liquid solution of xpb = 0.1. calculate the value of w for the system and calculate the activity of sn in the liquid solution of xsn = 0.5 at 500°c

Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln⁡〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J

Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT  X_Pb^2=(-4578 J)/(8.314 J/mol  x 773 K)  x 0.5 x 0.5= -0.718lnγ_Sn=exp⁡(-0.178)=0.386The activity of  Sn= γ_Sn  x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution  of xsn at 500 degree Celsius is 0.418

The answer is the first option just took the test