A) 52.3
B) 57.5
C) 55.8
D) 59.3
Correct Answer: C
Solution :
Sol. \[\underset{0.1596g}{\mathop{{{A}_{2}}{{O}_{3}}}}\,+\underset{0.006g}{\mathop{3{{H}_{2}}}}\,\xrightarrow{{}}2A+3{{H}_{2}}O\] \[\because \]0.006 g of\[{{H}_{2}}\]reduces 0.1596 g of\[{{A}_{2}}{{O}_{3}}\]. \[\therefore \] 6 g of\[{{H}_{2}}\]will reduce \[=\frac{0.1596}{0.006}\times 6g\,of\,{{A}_{2}}{{O}_{3}}\] \[=159.6g\,of\,{{A}_{2}}{{O}_{3}}\] Hence, molecular weight of\[{{A}_{2}}{{O}_{3}}=159.6\text{ }g\]and atomic weight of metal =55.8You need to login to perform this action.
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