31 grams of nacl is dissolved in 532 g of water. kb for water is .512 C/m . assume complete dissociation of the salt and ideal behavior of the solution. what is the boiling point elevation? answer in units of C.
----------------------------------
Here is what I did:
Solute = NaCl
Solvent = H2O
Solvent = kb = .512, n.b.p.= 100 deg celsius
deltaTb = (2)(.512)(0.53mol/0.532)
=1.02
100+1.02 = 101.02 deg Celsius
The final temperature that I gave which was 101.02 deg. celsius is the new boiling point of the solution. But is that what the question is asking for? It says it wants the boiling point ELEVATION, so does that mean the answer is 1.02 deg. celsius? Please clear this up for me..
----------------------------------
Here is what I did:
Solute = NaCl
Solvent = H2O
Solvent = kb = .512, n.b.p.= 100 deg celsius
deltaTb = (2)(.512)(0.53mol/0.532)
=1.02
100+1.02 = 101.02 deg Celsius
The final temperature that I gave which was 101.02 deg. celsius is the new boiling point of the solution. But is that what the question is asking for? It says it wants the boiling point ELEVATION, so does that mean the answer is 1.02 deg. celsius? Please clear this up for me..