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Dear student,
Let,
The mass of the organic compound taken be = W g
Volume of air displaced = V1 mL
Atmospheric pressure = P mm of Hg
Room temperature, T1 = toC = (t + 273)K
Aqueous tension at toC = p mm of Hg
Pressure of the dry air, P1 = (P-p) mm Hg
The volume of the displaced air is equal to the volume of vapors formed from the given mass of the compound. So, by using the mole concept, one can write,
V mL of vapour weigh = W g
Therefore,
VOLUME OCCUPIED BY 0.23 GM OF THE SUBSTANCE IS 112ML AT STP GIVEN
THEREFORE WT OF I MOLE OF THE SUBSTANCE IS WQUAL TO 22400 X .23 / 112
THAT IS MOLECULAR WEIGHT OF THE SUBSTANCE = 46
AND ITS VAPOUR DENSITY IS EQUAL TO 46/2 - 23?
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