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In a Victor Meyer's determination, the f...

In a Victor Meyer's determination, the following observations have been made:
Mass of compound `= 0.17g`
Volume of air collected `= 34.2 mL`
Temperature `= 15^(@)C`
Atmospheric pressure `= 750 mm`
Vapour pressure of water at `15^(@)C = 13mm`
Calculate the vapour density and molecular mass of the compound.

Text Solution

Verified by Experts

Given
`{:(V_(1)=34.2mL,V_(2)=?),(P_(1)=(750-13)=737mm,P_(2)=760mm),(T_(1)=(15+273)=288K,T_(2)=273K):}}NTP "conditions"`
By gas equation,
`V_(2)=(737 xx 34.2)/(288) xx (273)/(760) = 31.4376 mL`
Vapour density `=(W)/(V_(2) xx 0.00009) = (0.17)/(31.4376 xx 0.00009) = 60.08`
Mol.mass `= 2xx` Vapour density` = 2xx 60.08 = 120.16`
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