Question

See page 295 A common anesthesia gas is Halothane, with the formula C2HCIBrF3 and the structure is shown in the figure. Liquid Halothane has a density of 1.87 g/mL, and boils at 50.2°C and 1.00 atm.

theres 2 parts to the question. only answer it if u can answer both part 1 and 2

Answer 1

Part-1:

density = 1.87 g/mL; volume = 10 mL ==> mass = 10 mL x 1.87 g/mL = 18.7 g

moles = 18.7 g / 197.3815 g/mol = 0.09474038853 mol

PV = nRT ==> V = nRT/P = [ 0.09474038853 mol x 0.082057 L.atm/mol.K x (60 + 273.15)K] / 1.00 atm = 2.58994543351 L

Answer: 2.59 L

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Part-2:

Just change temperature as 55 ⁰C in above calculation to get volume

V = nRT/P = [ 0.09474038853 mol x 0.082057 L.atm/mol.K x (55 + 273.15)K] / 1.00 atm = 2.5510748732 L

this part is continuation of Part-1, so we have to stick with 18.7 g mass and also our volume calculations based on 18.7 g

density = mass / volume = 18.7 g / 2.5510748732 L = 7.33 g/L

Answer: 7.33 g/L

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Part 1 (1 point) See Periodic Table If Halothane behaved as an ideal gas, what volume would 10.0 mL of Halothane occupy at 60°C and 1.00 atm of pressure? L 2.59 (1 point) Part 2 What is the density of Halothane vapor at 55°C and 1.00 atm of pressure? g/L 7.33

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