$P = 1.13 \, atm$
$V = 10.0 \, L$
$T = 100.0^\circ C = 373.15 \, K$
$R = 0.0821 \, L \cdot atm \cdot mol^{-1} \cdot K^{-1}$
$n = \frac{PV}{RT} = \frac{(1.13 \, atm)(10.0 \, L)}{(0.0821 \, L \cdot atm \cdot mol^{-1} \cdot K^{-1})(373.15 \, K)} = 0.368 \, mol$
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