A) 32 g/mol
B) 37 g/mol
C) 42 g/mol
D) 98 g/mol
Correct Answer: A
Solution :
Let us suppose mass of A is x \[1\,\,\operatorname{mol}\,\,{{A}_{4}}{{O}_{6}}\,\,\equiv \,\,4\,mol\,A\] \[(4x+96)g {{A}_{4}}{{O}_{6}} \equiv \,\,4xg\,A\] \[\log \,\,{{A}_{4}}{{O}_{6}}=\frac{4x}{4x+96}\,\,\times \,\,\log A\] \[\frac{4x\times 10}{4x+96}=5.72\] \[\operatorname{A}= 32 g/mol\]
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