In determination by Victor Meyer method, 0.60 g of a volatile substance expelled 123 mL of air measured over water at 20^(@)C and 757.4 mm pressure. Find the molecular mass of the substance. Aqueous tension at 20^(@)C is 17.4 mm.

In determination by Victor Meyer method, 0.60 g of a volatile substanc

1.	In determination by Victor Meyer method, 0.60 g of a volatile substance expelled 123 mL of air measured over water at 20^(@)C and 757.4 mm pressure. Find the molecular mass of the substance. Aqueous tension at 20^(@)C is 17.4 mm.
Answer» SOLUTION :MASS of the organic substance `=0.60 g` Volume of the displaced air `=123.0 mL` Temperature at which air is collected `=20^(@)C=(20+273)=293 K` Pressure at which air is collected `=757.4 mm` Aqueous tension at `293 K=17.4 mm` Pressure of the dry air `=757.4-17.4=740 mm` STEP I. Calculationof the volume of the air displaced (or vapours of the substance) at N.T.P. `{:("EXPERIMENTAL conditions","N.T.P. conditions"),(V_(1)=123.0 mL,V_(2)=?),(P_(1)=740 mm,P_(2)=760 mm),(T_(1)=293 K,T_(2)=273 K):}` By applying gas equation : `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` or `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((740mm)xx(123.0 mL)xx(273 K))/((293K)xx(760 mm))=111.6 mL` Step II. Calculation of the molecular mass. 111.6 mL of vapours are obtained from subtance = 0.60 g 22400 mL of vapours are obtained from substance `=((0.60g)xx(22400 mL))/((111.6 mL))=120.4 g` `:.` Molecular mass of the substance `=120.4 g mol^(-1)`.