Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
In Victor meyers experiment, 0.23gram of a volatile solute displaced air which measures 112ml at 1 atm and 273k. Calculate vapour density and molecular weight of a substance
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,
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?
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?
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -