Question

an unknown gas at 51.1°c and 1.00 atm has a molar mass of 30.07 g/mol. assuming ideal behavior, what is the density of the gas?
density: g/l

Explanation:

Step1: Convert temperature to Kelvin

The ideal gas law is \(PV = nRT\). We know that \(n=\frac{m}{M}\) (where \(m\) is mass and \(M\) is molar mass), and density \(
ho=\frac{m}{V}\).
First, convert \(T = 51.1^{\circ}C\) to Kelvin. \(T=(51.1 + 273.15)K=324.25K\).
The ideal gas law can be rewritten as \(PV=\frac{m}{M}RT\), and then \(
ho=\frac{m}{V}=\frac{PM}{RT}\).

Step2: Substitute values into the formula

Given \(P = 1.00atm\), \(M=30.07g/mol\), \(R = 0.0821L\cdot atm/(mol\cdot K)\), and \(T = 324.25K\).
Substitute these values into \(
ho=\frac{PM}{RT}\):
\(
ho=\frac{1.00atm\times30.07g/mol}{0.0821L\cdot atm/(mol\cdot K)\times324.25K}\)

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Answer:

\(1.13g/L\)