# 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 $×$10-3 L
again from Dalton's law ;  Pmoist = Pdry + Aq. tension
Pdry   =   737.6 mm = 737.6 /760 = 0.97 atm
now from ideal gas eq.  PV = W/M  RT
then                             0.97 $×$ 45 $×$10-3  = 0.22 / M  $×$ 0.0821$×$   293
M     = 132.25 gm/mol
Regards

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