Here, we are given : V=18 litres, `T=27^(@)C`=27+273 K=300K
`P=70 cm=70//76" atm", "R"=0.0821" litre atm "K^(-1)mol^(-1)`
Using the ideal gas equation, PV=nRT
we have `" " n=(PV)/(RT)=((70//76" atm")(18 L))/((0.0821" L atm mol"^(-1))(300" K"))=0.67" mole"`
Further, we know that `n=("Mass)/("Molecular mass")=(m)/(M)" ":. M=(m)/(n)=(1.350)/(0.67)=2.015" u"`(Molar mass=mass `"mol"^(-1)`)