71It's an experiment to determine the molecular weight using the depression in vapour pressure thing.
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A current of dry air was bubbled through a bulb containing 26.6 g of an organic substance in 200 gms of water, then through a bulb at the same temperature containing pure water and finally through a tube containing fused calcium chloride. The loss in weight of water bulb is 0.0870 gms and gains in weight of CaCl2 tube is 2.036 gm. Calculate the molecular weight of the organic substance in solution.
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71
1When 2 gm of a non-volatile Hydrocarbon containing 94.4% carbon is dissolved in 100 gm of benzine. the V.P. of benzine at 20°c is loward from 74.66mm to 74.01mm. Calculate the molecular formula of the Hydrocarbon.
1we know,
gain in wt of CaCl2=loss in wt. of soln+loss in wt. of pure water
by this we get,
2.036=0.087+x...........x=loss in wt of soln
x=1.949
now according to raoults law,
m=(w.M.P0)/∂P.W
where
m=molecular wt.of solute
w=wt. of solute in grams in solution
W=wt. of solvent in grams
M=molecular wt. of solvent
∂P=loss in wt of pure water
P0=gain in wt. of CaCl2
now substituting the values
m=26.6*18*2.036/0.087*200
m=56.0210
i hope u might have understood
