1.

0.1693 gm of a volatile substance when vapourised displaced 58.9 cm of air measured at `27^@C` and 746 mm pressure. Calculate the molecular mass of the substance. (Aqueous tenstion at `27^@C=26.7mmHg`.)

Answer» To convert the volume at experimental conditions to volume at STP.
`{:("Experimental condition", "At STP"),(P_1=746-26.7=719.3mm,P_2=760mm),(V_1=58.9ml,V_2=?),(T_1=273+27=300K,T_2=273K):}`
Substituting these values in the gas equation,
`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`, we get
`(719.3mmxx58.9ml)/(300K)=(760mmxxV_2ml)/(273K)`
`V_2=(719.3mmxx58.9mlxx273K)/(300Kxx760mm)=50.73ml`
molecular mass `=("mass of substance"xx22400)/("volume of displaced air at "STP)`
`=(0.1693xx22400)/(50.73)=74.75gm`


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