1.

An oleum sample labelled as 104.5% in 10 g of this sample 90 mg water is added then which is/are correct for resulting solution.

Answer»

Solution contain 10.09 g `H_2SO_4`
Solution contain 15.86% FREE `SO_3`
Solution contain 8.49 g `H_2SO_4`
Solution contain 20% free `SO_3`

Solution :104.5 % means
100 g oleum sample requir 4.5 g `H_2O` to PRODUCED 104.5 g `H_2SO_4`
So, for 10 g sample water required =0.45 g
`SO_3+underset(0.45 g)(H_2O)toH_2SO_4`
So, free `SO_3=80/18xx0.45=2 g`
Now, 0.09 g `H_2O` is ADDED which react with `SO_3`
`underset(2G)(SO_3)+underset(0.09 g)(H_2O)toH_2SO_4`
MOLE `1/40 1/200(L.R.)`
`{:("After reaction",0.02,0,0.005),("wt.",1.6g,0,0.49 g):}`
Now wt. of `H_2SO_4=8+0.49=8.49 g`
% of free `SO_3=(1.6xx100)/(8.49+1.6)=15.86%`


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