Check my thinking on this carefully.
I looked up the density of liquid butane. It is 0.573 g/mL and since -22C is well below the boiling point the volume occupied by the liquid is v = m/d = 0.65/0.573 = 1.13 mL for the first part.
For the second part, I am assuming the jar was opened, the butane inserted (at -22 C) then the jar was closed with the pressure of 1 atm.
So the volume of the container is 250-1.13 = 248.9 mL. How much of the butane can evaporate when the T is raised to 25 C? That is
n = PV/RT = 1 atm x 0.2489/0.08206*298 = 0.01. You had 0.011 mols initially. The difference is 0.011-0.01 = 0.001 left as a liquid. That translates to
0.001 x 58 = approx 0.06 g left as a liquid or about 0.06/0.573 or about 0.1 mL. Check my thinking.
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mole and a normal boiling point of -0.4 C. A 250 mL sealed flask contains 0.65 g of butane at -22 C.
How much butane is present as a liquid at -22C?
If the butane is warmed to 25 C, how much liquid butane is present?
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