A) 75%
B) 30.6%
C) 25%
D) 69.4%
Correct Answer: A
Solution :
\[\text{1}:\text{2}\] \[\text{2}:\text{1}\]Number of moles of \[{{L}_{1}}\]produced\[{{L}_{2}}\] \[{{L}_{2}}\]Number of moles of CaC03produced = 0.01 Number of moles \[{{L}_{2}}\]taken = 0.01 \[\sqrt{{{L}_{1}}}\] Mass of \[{{L}_{1}}\] (Molar mass of \[I\]) \[\frac{1}{2}\] Mass of \[\text{125J}\] \[\text{1}.\text{25J}\]Percentage of \[\text{P}{{\text{T}}^{^{\text{2}}}}=\]You need to login to perform this action.
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