a) s20=372.0 g/L (20°C)
s100=740.8 g/L (100 °C)
M(NH4Cl)=53 g/mol
S20=s20/M=372/53=7 mol/L
S100=s100/M=740.8/53=14 mol/L
b) s20=218.0 g/L (20 °C)
s100=436.0 g/L (100 °C)
M(Na2CO3)=106 g/mol
S20=s20/M=218/106=2.06 mol/L
S100=s100/M=436/106=4.11 mol/L
Comments
Leave a comment