Example 9.51b: Backward fading, part 3
How many grams of gas are present in 221 mL of Ar at 0.23 torr and –54 °C.
Plan 1) \(n_{\mathrm{\ce{Ar}}}\) 2) \(m_{\mathrm{\ce{Ar}}}\)
\(V_{\mathrm{\ce{Ar}}}\) \(= 221.\ \mathrm{mL}\)
\(\ \ \ =0.221\ \mathrm{L}\)
\(P\) \(= 0.23\ \mathrm{torr}\)
\(\ \ \ =3.0\times 10^{-4}\ \mathrm{atm}\)
\(T\) \(= -54.\ \mathrm{°aC}\)
\(\ \ \ =219.\ \mathrm{K}\)
What did we just do before starting any calculations?
\(n_{\mathrm{\ce{Ar}}}\) = ?
\(m_{\mathrm{\ce{Ar}}}\) = ?
Think about it...