A) 32 amu
B) 64 amu
C) 67 amu
D) 98 amu
Correct Answer: A
Solution :
Molecular weight of\[{{X}_{4}}{{O}_{6}}=(4a+96)\] \[\because \,\,(4a+96)g\,\,{{X}_{4}}{{O}_{6}}\]contains\[X=4ag\] \[\therefore \]\[10\,\,g{{X}_{4}}{{O}_{6}}\]contains\[X=\frac{4a\times 10}{4a+96}\] \[5.72=\frac{4a\times 10}{4a+96}\] \[22.88a+549.12=40a\] \[17.12a=549.12\] \[a=32.07\,\,amu\]You need to login to perform this action.
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