Question

A 12.4 mL sample of liquid butane (density = 0.573 g/mL is evaporated in an otherwise empty container at a temperature of 30.3 C. The pressure in the containerfollowing evaporation is 885 torr. What is the volume of the container?

Answer 1

First, we need to find the mass of evaporated butane by multiplying its liquid volume by its liquid density:

(12.4 mL)(0.573 g/mL) = 7.1052 g

Next, we need to convert 885 torr to atmospheres:

(1 atm) : (760 torr) = (X) : (885 torr)
X = [(1 atm)(885 torr)] / (760 torr)
X = 1.1644736 atm

Next, we need to convert 29.2°C to Kelvins:

30.3°C + 273.15 = 303.45 K

Next, w can use the following equation to solve for the volume of the gas container:

(P)(V) = (n)(R)(T)
or
V = [(n)(R)(T)] / (P)
n = 7.1052 g / 58.12g/mol
= 0.122 moles
V = [(0.122 mol)(0.0821L·atm/mol·K)(303.45 K)] / (1.1644736 atm)
V = 2.6101200 L or 2.610 L rounded to four significant figures

Answer 2

: The volume of the container is about 2.610 Liters.