A) 0.63 gm
B) 0.1575 gm
C) 0.126 gm
D) 0.875 gm
Correct Answer: C
Solution :
| [c] Per mole of given acid \[{{C}_{2}}O_{4}^{2-}\] present = 1 mole |
| \[\therefore \] \[{{C}_{2}}O_{4}^{2-}\to C{{O}_{2}}+2e\] |
| \[\therefore \] x factor for \[{{H}_{2}}{{C}_{2}}{{O}_{4}},2{{H}_{2}}O\] as a reducing agent = 2 |
| \[\therefore \] equiv. wt. \[=\frac{250}{100}\times {{10}^{-3}}=63\] |
| 250 ml of seminormal solution \[=\frac{250}{100}\times {{10}^{-3}}\] equiv. |
| \[\therefore \] wt. of \[{{H}_{2}}{{C}_{2}}{{O}_{4}},2{{H}_{2}}O\] required |
| \[=\frac{250}{100}\times {{10}^{-3}}\times 63gm=0.1575gm\] |
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