A) \[52.3\]
B) \[57.5\]
C) \[55.8\]
D) \[59.3\]
Correct Answer: C
Solution :
\[\underset{0.1596\text{ }g}{\mathop{{{A}_{2}}{{O}_{3}}}}\,+\underset{0.006\text{ }g}{\mathop{3{{H}_{2}}}}\,\xrightarrow[{}]{{}}2A+3{{H}_{2}}O\] 0.006 g of\[{{H}_{2}}\]reduces 0.1596 g of\[{{A}_{2}}{{O}_{3}}\] 6 g of\[{{H}_{2}}\]will reduces \[=\frac{0.01596\times 6}{0.006}=159.6\,g\,{{A}_{2}}{{O}_{3}}\] Hence, molecular weight of\[{{A}_{2}}{{O}_{3}}=159.6\text{ }g\] Let molecular weight of \[A=x\] \[\therefore \] \[2x\times 3\times 16=159.6\] or \[2x=159.6-48\] or \[2x=111.6\] or \[x=55.8\]
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