I have 20 ml 68% nitric acid in a 50 ml closed vessel which vents when
100 psi is exceeded. What is the maximum temperature this vessel can
see without venting the boiling acid?

Vol. of the HNO3 = 20ml = 0.02L
Concentration of the HNO3 = 68%
Vol. of the vessel = 50ml = 0.05L
Pressure = 100psi = 6.80459639 atm
Temperature = ?

1.
Concentration of the HNO3 = (# of mole of the HNO3 / Vol. of the HNO3) * 100%
68% = (# of mole of the HNO3 / 0.02) * 100%
# of mole of the HNO3 = 0.0136 (moles)

2.
Use Ideal Gas Law:
PV = nRT

P = 6.80 atm (see above)
V = 0.0500 L (see above)
R = 0.0821   (Constant)
n = 0.0136 moles (from part 1)
T = unknown

Solve "T":

T = 305K = 32 oC

That means if the temperature is under 32 oC,
this vessel can see without venting the boiling acid.

This is not quite right. To use the ideal gas law you are assuming
that the 20ml of liquid is in the vapour phase. It is actually in the
liquid phase, and some will evaporate off given a specific temperature
and pressure. So there are two unknowns, the temperature, and the No
of moles of the 20 ml that will be in the vapour phase at the
particular temperature and pressure.

Also, nitric acid does not obey the Ideal Gas law, infact it deviates
significantly from it. I will accept solutions that make the Ideal Gas
assumption if all other aspects of the solution are correct.

To explain what I am actually doing, I have 20 ml of the acid in a 50
ml vessel and I am going to keep heating it until liquid evaporates
off and the pressure reaches 100 psi. Not necessarily all the liquid
will have evaporated before this pressure is reached.