In a Victor Meyer's determination of molecular mass, 0.1015 g of an organic substance displaced 27.96 mL of air at 15°C and 766 mm pressure. Calculate the molecular mass of the substance (Aqueous tension at 15°C = 16 mm).

In a Victor Meyer's determination of molecular mass, 0.1015 g of an or

1.	In a Victor Meyer's determination of molecular mass, 0.1015 g of an organic substance displaced 27.96 mL of air at 15°C and 766 mm pressure. Calculate the molecular mass of the substance (Aqueous tension at 15°C = 16 mm).
Answer» Solution :In the present case, mass of the substance = 0.1015 g Volume of moist air = 27.96 mL Room temperature = 273 + 15 = 288 K Barometric pressure = 766 MM `P_(1) = 766-16 = 750 mm, P_(2) = 760 mm` `V_(1) = 27.96 mL, V_(2)`= ? `T_(1) = 288 K, T_(2) = 273 K` According to the gas equation, `(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)` `V_(2) = (P_(1)V_(1))/T_(1) xx T_(2)/P_(2) = (750 xx 27.96)/288 xx 273/760 = 26.16 mL` Thus, 0.1015 g of the given compound displaces 26.16 mL of dry air at S.T.R This is equal to the volume of vapour formed. `THEREFORE 26.16` mL of vapour at S.T.R weigh = 0.1015 g `=0.1015/26.16 xx 22400 = 86.9 g` Hence, the MOLECULAR mass of the given substance is 86.9