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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. |
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Answer» SOLUTION :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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