A) \[1.0\]
B) \[12.7\]
C) \[13.0\]
D) \[1.30\]
Correct Answer: C
Solution :
\[2NaCl+2{{H}_{2}}O\xrightarrow{electrolysis}{{H}_{2}}+C{{l}_{2}}\] \[+2NaOH\] Weight of \[NaCl\]I present in \[0.5L=0.5mol\] Charge \[=965\times 5=4825C\] \[\therefore \] Number of moles decomposed \[=\frac{1\times 4825}{96500}=0.05mol\] \[\therefore \] Number of moles of \[NaOH\] formed is also 0.05. \[\therefore \] \[Molarity=\frac{0.05\times 1000}{500}=0.1\] \[pOH=1,pH=13\]
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