Question

calculate to three significant digits the density of carbon monoxide gas at exactly 35 °c and exactly 1 atm. you can assume carbon monoxide gas behaves as an ideal gas under these conditions.

Explanation:

Step1: Recall the 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 the molar mass. Substituting $n$ into the ideal gas law: $PV=\frac{m}{M}RT$, then rearranging for density $
ho=\frac{PM}{RT}$.

Molar mass of CO: $M = 12.01 + 16.00 = 28.01\ g/mol$. Pressure $P = 1\ atm$, Temperature $T = 35^\circ C + 273.15 = 308.15\ K$, Gas constant $R = 0.0821\ \frac{L\cdot atm}{mol\cdot K}$.

Step2: Substitute values into the density formula

$
ho=\frac{PM}{RT}=\frac{1\ atm\times28.01\ g/mol}{0.0821\ \frac{L\cdot atm}{mol\cdot K}\times308.15\ K}$

First, calculate the denominator: $0.0821\times308.15\approx25.299$

Then, calculate the numerator: $1\times28.01 = 28.01$

Now, divide: $
ho=\frac{28.01}{25.299}\approx1.107\ g/L$

Step3: Round to three significant digits

Rounding $1.107$ to three significant digits gives $1.11\ g/L$.

Answer:

$1.11$