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?.

0.260 g
0.515 g
0.760 g
1.00 g

Answer 1

The answer to your question is: mass = 0.760 g

Step-by-step 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