Question

How many grams of pentane are contained in a vapor sample that occupies a volume of 0.285 L at 57°C and 1.0 atm?.

Answer 1

Answer:

The answer to your question is: mass = 0.760 g

Explanation:

Data

V = 0.285 L

T = 57°C = 330°K

P = 1 atm

R = 0.082 atm L/mol°K

mass = ?

Formula

                PV = nRT

                n = PV / RT

Substitution

                n = (1)(0.285) / (0.082)(330)

                n = 0.285 / 27.06

                n = 0.011

mW Pentane = (12 x 5) + (12 x 1) = 72 g

                                 72 g of pentane ----------------- 1 mol

                                   x                        ----------------- 0.0105 mol

                                  x = (0,0105 x 72) / 1

                                  x = 0.756 g ≈ 0.760 g