A) 5.4 g
B) 10.8 g
C) 54.0 g
D) 108.0 g
Correct Answer: D
Solution :
| [d] Since, 22400 mL volume is occupied by 1 mole of \[{{O}_{2}}\] at STP. |
| Thus, 5600 mL \[{{O}_{2}}\] means = \[\frac{5600}{22400}\text{mol}{{O}_{2}}\] |
| \[=\frac{1}{4}\text{mol}\,{{O}_{2}}\] |
| \[\therefore \] Weight of \[{{O}_{2}}\frac{1}{4}\times 32=8g\] |
| According to problem, |
| Equivalents of Ag = Equivalents of \[{{O}_{2}}\] |
| =\[\frac{\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ Weight of Ag}}{\text{Equivalent weight of Ag}}\] |
| =\[\frac{{{\text{W}}_{O}}_{_{\text{2}}}}{\text{Equivalent}\,\text{weight}\,\text{of}\,{{\text{O}}_{\text{2}}}}\] |
| \[\frac{{{W}_{Ag}}}{\frac{{{M}_{Ag}}}{VF}}=\frac{{{W}_{{{O}_{2}}}}}{\frac{{{M}_{{{O}_{2}}}}}{VF}}\] |
| \[\therefore \] \[\frac{{{W}_{Ag}}}{108}\times \frac{8}{32}\times 4\] |
| \[[\because \,2{{H}_{2}}O\to {{O}_{2}}+4{{H}^{+}}+4{{e}^{-}}]\] |
| Þ \[{{W}_{Ag}}=108g\] |
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