Question

What is the volume of 0.190 g of C2H2F4 vapor at 0.606 atm and 37.5◦C?
Answer in units of L.

Answer 1

pV=nRT (Mendeley-Claveiron equation) V=nRT/p

R=const=8.314

M(C₂H₂F₄)=12*2+1*2+19*4=102 (g/mole)

n(C2H2F4)=m/M=0.19/102=0,00186 (mole)

T=37.5◦C +273 = 310.5 K

p=0,606 atm *101325= 61.40295 Pa

V=0.00186*8.314*310.5/61.40595=0,078 L

answer: 0.078 L