A) 1 : 2
B) 2 : 1
C) 4 : 5
D) 5 : 4
Correct Answer: C
Solution :
eQ. of \[N{{a}_{2}}{{C}_{2}}{{O}_{4}}\] + eQ. of \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\] = eQ. of \[KMn{{O}_{4}}=V\,\,\times \,\,0.1\,\,\times \,\,5\] |
\[2+2=0.5V;\,\] \[V=8L\] |
eQ. of \[{{H}_{2}}{{C}_{2}}{{O}_{4}}=eQ.\]of \[NaOH\] \[\Rightarrow \,\,\,1\,\,\times \,\,2=0.2\,\,\times \,\,{{V}_{2}};{{V}_{2}}=10L\] |
Hence, \[{{V}_{1}}:{{V}_{2}}=8:10=4:5\] |
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