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+//Example11.1//
+Ni=0.55;//wt % // steel and nominal alloy content
+Cr=0.50;//wt % //steel and nominal alloy content
+Mo=0.20;//wt % //steel and nominal alloy content
+C=0.30;//wt %//steel and nominal alloy content
+a=100-(Ni+Cr+Mo+C)
+mprintf("a = %f g Fe ",a)
+a1=55.85;//g /mol // atomic mass of iron
+b=0.6023*10^24;//atoms/ mol //Avagardo's constant
+NFe=(a/a1)*b
+mprintf("\nNFe = %e atoms",NFe)
+//similarly
+c=58.71;//g/mol // atomic mass of nickel
+Nni=(Ni/c)*b
+mprintf("\nNni = %e atoms",Nni)
+d=52.00;//g/mol // atomic mass of chromium
+NCr=(Cr/d)*b
+mprintf("\nNCr = %e atoms",NCr)
+e=95.94;//g/mol //atomic mass of Molybdenum
+NMo=(Mo/e)*b
+mprintf("\nNMo = %e atoms",NMo)
+f=12.01;//g/mol //atomic mass of Carbon
+NC=(C/f)*b
+mprintf("\nNC = %e atoms",NC)
+//so in a 100-g there shold be
+Ntotal=NFe+Nni+NCr+NMo+NC
+mprintf("\nNtotal = %e atoms",Ntotal)
+//The atomic fraction of each alloying element is then
+XNi=Nni/Ntotal
+mprintf("\nXNi = %e ",XNi)
+XCr=NCr/Ntotal
+mprintf("\nXCr = %e",XCr)
+XMo=NMo/Ntotal
+mprintf("\nXMo = %e",XMo)
+Xc=NC/Ntotal
+mprintf("\nXc = %e",Xc)
+XNi=5.19*10^-3;//atoms
+XCr=5.32*10^-3;//atoms
+XMo=1.16*10^-3;//atoms
+//which for a 100000 atom alloy gives
+h=10^5;//atoms //given
+NNi=XNi*h
+mprintf("\nNNi = %i atoms",NNi)
+NCr=XCr*h
+mprintf("\nNCr = %i atoms",NCr)
+NMo=XMo*h
+mprintf("\nNMo = %i atoms",NMo)
+Nc=Xc*h
+mprintf("\nNc = %i atoms",Nc)
+