What is the boiling point when 14.09 g of aspirin (a molecular compound with a molar...
What is the boiling point when 14.09 g of aspirin (a molecular compound with a molar mass = 176.60 g/mol) is dissolved in 199.10 g of benzene (solvent)?
Homework Answers
Boiling point elevation= kb*m, m= molality = moles of solute/ kg of solvent
moles of aspirin = 14.09/176.6=0.0798. mass of benzene= 199.10g= 0.199 kg
Molality= 0.0798/0.199=0.40
Kb = 2.53 degc/m
Boiling point elevation= 2.53*0.4=1.012 deg.c
Boiling point of Benzene= 80.1 deg.c, actual boiling point when aspirin is added= 80.1+1.012=81.112 deg.c
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