Answer:
the answer is C hope I helped
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
Ionic substances formed from complete transfer of electrons from one atom to another causing electrostatic attraction
Answer:
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