A) \[125mL\]
B) \[25mL\]
C) \[12.5mL\]
D) \[37.5mL\]
Correct Answer: C
Solution :
: \[2KMn{{O}_{4}}+5{{(COOH)}_{2}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}\]\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+10C{{O}_{2}}+8{{H}_{2}}O\] 2 moles of \[KMn{{O}_{4}}=5\] moles of \[{{(COOH)}_{2}}\] \[\therefore \] \[20\times 0.025\times {{10}^{-3}}\] moles of \[KMn{{O}_{4}}\] \[\equiv \frac{5}{2}\times 20\times .025\times {{10}^{-3}}\] \[=1.25\times {{10}^{-3}}\]moles of \[{{(COOH)}_{2}}\] \[0.1\] mole of oxalic acid are present in 1000 mL \[\therefore \] \[1.25\times {{10}^{-3}}\] mole will be present in \[=\frac{1000}{0.1}\times 1.25\times {{10}^{-3}}=12.5mL\]
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