if 0.22 gram of a substance when vaporized displaced 45 cm cube of air measured over Water at 293 Kelvin and 755mm pressure and if vapour pressure of H2O is 17.4mm then a molecular weight of substance will be

Dear student ,
as we know that ; vol. of vapours of solid substance = vol. of air displaced = 45 cm³ = 45 $\times$10^-3 L
again from Dalton's law ;  P_moist = P_dry + Aq. tension 
                                            P_{dry   =}   737.6 mm = 737.6 /760 = 0.97 atm
now from ideal gas eq.  PV = W/M  RT
then                             0.97 $\times$ 45 $\times$10^-3  = 0.22 / M  $\times$ 0.0821$\times$   293 
                                                        M     = 132.25 gm/mol
Regards