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carbon monoxide is a gas at 0 °c and a pressure of 1.03 × 10⁵ pa. it is a diatomic gas, each of its molecules consisting of one carbon atom (atomic mass = 12.0 u) and one oxygen atom (atomic mass = 16.0 u). assuming that carbon monoxide is an ideal gas, calculate its density ρ.
ρ =
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Explanation:

Step1: Recall ideal gas law and density formula

The ideal gas law is $PV = nRT$, and density $
ho=\frac{m}{V}$, where $n=\frac{m}{M}$ ( $M$ is molar mass). Substituting $n$ into ideal gas law gives $PV=\frac{m}{M}RT$, then rearranging for $
ho$: $
ho=\frac{PM}{RT}$.

Step2: Calculate molar mass of CO

Molar mass of C is $12.0\ \text{g/mol}$, O is $16.0\ \text{g/mol}$. So $M = 12.0 + 16.0 = 28.0\ \text{g/mol}=0.028\ \text{kg/mol}$.

Step3: Identify values for P, R, T

$P = 1.03\times 10^{5}\ \text{Pa}$, $R = 8.314\ \text{J/(mol·K)}$, $T = 0^\circ\text{C}=273.15\ \text{K}$.

Step4: Substitute into density formula

$
ho=\frac{1.03\times 10^{5}\times0.028}{8.314\times273.15}$.

Calculate numerator: $1.03\times 10^{5}\times0.028 = 2884$.

Denominator: $8.314\times273.15\approx2271.0$.

Then $
ho=\frac{2884}{2271.0}\approx1.27\ \text{kg/m}^3$ (or in g/L, since $1\ \text{kg/m}^3 = 0.001\ \text{g/cm}^3 = 1\ \text{g/L}$, wait no: $1\ \text{kg/m}^3=1\ \text{g/L}$? Wait no, $1\ \text{m}^3 = 1000\ \text{L}$, so $1\ \text{kg/m}^3=\frac{1000\ \text{g}}{1000\ \text{L}} = 1\ \text{g/L}$. Wait, but let's check units: $P$ in Pa (N/m²), $M$ in kg/mol, $R$ in J/(mol·K) (which is m²·kg/(s²·mol·K)), $T$ in K. So units: $\frac{(\text{N/m}^2)(\text{kg/mol})}{(\text{m}^2·\text{kg}/(\text{s}^2·\text{mol}·\text{K}))(\text{K})}=\frac{\text{kg}/\text{m}^3}{1}$, so units are kg/m³.

Wait, let's recalculate:

$P = 1.03\times 10^{5}\ \text{Pa}$, $M = 0.028\ \text{kg/mol}$, $R = 8.314\ \text{J/(mol·K)}$, $T = 273\ \text{K}$ (approx).

$
ho=\frac{1.03\times 10^{5}\times0.028}{8.314\times273}$

$1.03\times10^{5}\times0.028 = 1.03\times2800 = 2884$

$8.314\times273\approx8.314\times273 = 2270.722$

$
ho=\frac{2884}{2270.722}\approx1.27\ \text{kg/m}^3$ (or $1.27\ \text{g/L}$ since $1\ \text{kg/m}^3 = 1\ \text{g/L}$? Wait no, $1\ \text{kg/m}^3 = 0.001\ \text{g/cm}^3 = 1\ \text{g/L}$? Wait, $1\ \text{m}^3 = 1000\ \text{L}$, so $1\ \text{kg/m}^3=\frac{1000\ \text{g}}{1000\ \text{L}} = 1\ \text{g/L}$. So $1.27\ \text{kg/m}^3 = 1.27\ \text{g/L}$. But let's check with exact $T = 273.15\ \text{K}$:

$8.314\times273.15 = 8.314\times273 + 8.314\times0.15 = 2270.722 + 1.2471 = 2271.9691$

$
ho=\frac{2884}{2271.9691}\approx1.27\ \text{kg/m}^3$ (or $1.27\ \text{g/L}$).

Answer:

$\boxed{1.27\ \text{kg/m}^3}$ (or $\boxed{1.27\ \text{g/L}}$)