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clc
clear
printf("example 2.3 page number 72\n\n")
//to find equivalent mass of metal
P=104.3 //total pressure in KPa
pH2O=2.3 //in KPa
pH2=P-pH2O; //in KPa
VH2=209*pH2*273/(293*101.3)
printf("volume of hydrogen obtained = %f ml\n\n",VH2)
//calculating amount of metal having 11.2l of hydrogen
m=350/196.08*11.2 //mass of metal in grams
printf("mass of metal equivalent to 11.2 litre/mol of hydrogen = %f gm",m)
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