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carbon monoxide is a gas at 0 °c and a pressure of 1.40 × 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 \(
ho=\frac{m}{V} \). Also, \( n = \frac{m}{M} \) where \( M \) is molar mass. Substituting \( n \) into ideal gas law: \( PV=\frac{m}{M}RT \), then rearranging for \(
ho=\frac{m}{V}=\frac{PM}{RT} \).

Step2: Calculate molar mass of CO

Carbon atomic mass: \( 12.0\ u \), Oxygen: \( 16.0\ u \). Molar mass \( M=(12.0 + 16.0)\ g/mol=28.0\ g/mol = 0.028\ kg/mol \).

Step3: Identify values for \( P, R, T \)

\( P = 1.40\times10^{5}\ Pa \), \( R = 8.314\ J/(mol\cdot K) \), \( T = 0^\circ C=273.15\ K \).

Step4: Substitute into density formula

\(
ho=\frac{PM}{RT}=\frac{(1.40\times10^{5}\ Pa)(0.028\ kg/mol)}{(8.314\ J/(mol\cdot K))(273.15\ K)} \).

Calculate numerator: \( 1.40\times10^{5}\times0.028 = 3920 \).

Denominator: \( 8.314\times273.15\approx2271.5 \).

\(
ho=\frac{3920}{2271.5}\approx1.726\ kg/m^3 \).

Answer:

\( \approx 1.73\ kg/m^3 \) (or more precise \( 1.726\ kg/m^3 \))