## Chemistry: Molecular Approach (4th Edition)

$Ar$ : 39.95 g/mol 1. Using the molar mass as a conversion factor, find the amount in moles: $$12.5 \space g \times \frac{1 \space mole}{ 39.95 \space g} = 0.313 \space mole$$ 2. According to the Ideal Gas Law: $$V = \frac{nRT}{P} = \frac{( 0.313 \space mol)( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 322 \space K)}{ 1.05 \space atm }$$ $$V = 7.88 \space L$$ 3. If the sample were 12.5 g of He: $He$ : 4.003 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$12.5 \space g \times \frac{1 \space mole}{ 4.003 \space g} = 3.12 \space moles$$ 4. According to the Ideal Gas Law: $$V = \frac{nRT}{P} = \frac{( 3.12 \space mol)( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 322 \space K)}{ 1.05 \space atm }$$ $$V = 78.5 \space L$$