Answer
0.193 atm
Work Step by Step
Volume $V=10.0\,L$
Number of moles $n=\frac{1.00\,g}{18.015\,g/mol}=0.0555\,mol$
Universal gas constant $R= 0.0821\,L\,atm\,mol^{-1}K^{-1}$
Temperature $T=(150.\,+273)K= 423\,K$
Recall that $PV=nRT$ (ideal gas law)
$\implies $ Pressure $P= \frac{nRT}{V}$
$=\frac{(0.0555\,mol)(0.0821\,L\,atm\,mol^{-1}K^{-1})(423\,K)}{10.0\,L}$
$=0.193\,atm$
