clc
//initialisation of variables
clear
L2= 4120 //cal
l= -108 //cal mole^-1
L21= -306 //cal mole^-1
n1= 55.5 //moles
n2= 1 //mole
//CALCULATIONS
Q= L21+L2
//RESULTS
printf ('differential heat of solution = %.f cal mole^-1',Q)