Question

Question #0f168

Answer 1

The density of the gas under the new conditions will be `"0.15 g/L"`.

Since density is defined as mass per unit of volume, you need to determine how much gas you have in the container. To do that, you need the number of moles present.

The key here is to realize that the number of moles remains constant - the only paramaters that change are pressure and temperature.

In other words, you'll have the same amount of gas in a larger volume, so you would expect the density to decrease.

Use the ideal gas law equation to solve for the number of moles of carbon monoxide present

`PV = nRT => n = (PV)/(RT)`

`n_"CO" = (1.13cancel("atm") * 0.48cancel("L"))/(0.082(cancel("L") * cancel("atm"))/("mol" * cancel("K")) * 349cancel("K")) = "0.01895 moles CO"`

Use carbon monoxide's molar mass to determine the mass of the gas

`m_"CO" = 0.01895cancel("moles") * 28.01"g"/cancel("mol") = "0.531 g"`

Now you have mass and volume, which means you can calculate density

`rho = m/V = "0.531 g"/"3.6 L" = "0.147 g/L"`

Rounded to two sig figs, the number of sig figs given for the two volumes, the answer will be

`rho = color(green)("0.15 g/L")`

SIDE NOTE The initial density of the gas was

`rho_"initial" = "0.531 g"/"0.49 L" = "1.1 g/L"`