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sample of impure Magnesium is reacted with diluted Sulphuric Acid to give the respective salt and Hydrogen.If 1 gram of impure sample gave 298.6 cc of Hydrogen at STP.Calculate the % purity of the sample (MAgnesium = 24,Hydrogen = 1)
Magnesium is reacted with diluted Sulphuric Acid gives the following reaction:
Moles of H2 at STP = (298.6)/(22.4 x 1000) = 0.0133
Mg + H2SO4 -------> MgSO4 + H2
0.0133
Thus, stoich. coeff. are equal hence moles of Mg are same as H2 .i.e = 0.0133
Now, Mg in gm = moles x molar mass of Mg
= 0.0133 x 24
=0.3192 gm
Now percentage of purity of the sample is = 0.3192/1 x 100
= 31.9 or 32 % (approx)
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