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## Q. 5.8

Calculate the density of carbon dioxide $(CO_{2}$) in grams per liter (g/L) at 0.990 atm and 55°C.

Strategy We need Equation (5.11) $d=\frac{m}{V}=\frac{P.M}{RT}$

to calculate gas density. Is sufficient information provided in the problem? What temperature unit should be used?

## Verified Solution

To use Equation (5.11), we convert temperature to kelvins (T = 273 + 55 = 328 K) and use 44.01 g for the molar mass of  $CO_{2}$; that is,

$d=\frac{P.M}{RT}$

$=\frac{(0.990 atm)(44.01 g/mol)}{(0.0821 L . atm/K . mol)(328 K)}=1.62 g/L$

Alternatively, we can solve for the density by writing

$density \frac{mass}{volume}$

Assuming that we have 1 mole of $CO_{2}$, the mass is 44.01 g. The volume of the gas can be obtained from the ideal gas equation

$V=\frac{nRT}{P}$

$=\frac{(1 mol)(0.0821 L . atm/K . mol)(328K)}{0.990 atm}$

=27.2  L

Therefore, the density of  $CO_{2}$  is given by

$\frac{44.01 g}{27.2 L}=1.62 g/L$

Comment In units of grams per milliliter, the gas density is $1.62 \times 10^{-3}$g/mL, which is a very small number. In comparison, the density of water is 1.0 g/mL and that of gold is 19.3 g/$cm^{3}$.