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#101. If 25.20 g of non-volatile, non-dissociating solute was dissolved in 600ml of H2O, the resulting solution boiled at 100.4C. i fthe Kb= 0.51C/m, what is the approximate molecular mass of the solute?
ans.
deltaT = Kb * m* i
.4 = .51m
.8 = m
.8= moles of solute/ kg solvent
.8 = moles/.600 ==> .5moles
MW= g/moles = 25.20/.5 = 50.4g/mole
I dont' understand how 600ml were kg solvent
does ml=kg?????? am lost........can some1help.??
ans.
deltaT = Kb * m* i
.4 = .51m
.8 = m
.8= moles of solute/ kg solvent
.8 = moles/.600 ==> .5moles
MW= g/moles = 25.20/.5 = 50.4g/mole
I dont' understand how 600ml were kg solvent
does ml=kg?????? am lost........can some1help.??