A) 83.9
B) 93.8
C) 98.3
D) 89.3
Correct Answer: D
Solution :
\[Mg+{{H}_{2}}S{{O}_{4}}\xrightarrow{\,}\,MgS{{O}_{4}}+{{H}_{2}}\uparrow \] 1 mol 1 mol = 24 g = 22.4 L 24 g of Mg (pure) \[\equiv \,22.4\,L\,of\,{{H}_{2(g)}}\] at STP at STP \[\therefore \] 2.4 g (pure) = 2.24 L of \[{{H}_{2}}\] at STP \[\therefore \] Percentage purity \[=\frac{2.00}{2.24}\times 100=89.3%\]
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