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Example 9.50: Backward fading part 1

How many grams of gas are present in 0.100 L of \(\ce{CO2}\) at 307 torr and 26 °C?

\(V_{\mathrm{\ce{CO2}}}\) \(= 0.100\ \mathrm{L}\)


\(P\) \(= 307.\ \mathrm{torr}\)

\(\ \ \ =0.404\ \mathrm{atm}\)


\(T\) \(= 26.\ \mathrm{°aC}\)

\(\ \ \ =299.\ \mathrm{K}\)


1. Use the ideal gas law to obtain the chemical amount of \(\ce{CO2:}\)

\(R\) \(= 8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)

\(\ \ \ =0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}}\)


\(n_{\mathrm{\ce{CO2}}}\) \(= \dfrac{P \cdot V_{\mathrm{\ce{CO2}}}}{R \cdot T}\)

\(\ \ \ =\dfrac{0.404\ \mathrm{atm} \cdot 0.100\ \mathrm{L}}{0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}} \cdot 299.\ \mathrm{K}}\)

\(\ \ \ =\dfrac{0.04039\ \mathrm{L}\ \mathrm{atm}}{24.546\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}}}\)

\(\ \ \ =1.65\times 10^{-3}\ \mathrm{mol}\)


2. Convert chemical amount of a pure compound to mass:

\(m_{\mathrm{\ce{CO2}}}\) \(= n_{\mathrm{\ce{CO2}}} \cdot M_{\mathrm{\ce{CO2}}}\)

\(\ \ \ =1.65\times 10^{-3}\ \mathrm{mol} \cdot 44.009\ \frac{\mathrm{g}}{\mathrm{mol}}\)

\(\ \ \ =0.0724\ \mathrm{g}\)


a) Tell me about the first step. How do we know we can use the ideal gas law?
b) Tell me about the second step. What do we need to know to convert between chemical amount and mass?