A) 75
B) 31.5
C) 40.2
D) 25
Correct Answer: A
Solution :
\[\underset{0.56\,\,g}{\mathop{CaO}}\,\xleftarrow{{}}\underset{1\,\,g}{\mathop{CaC{{O}_{3}}}}\,\xleftarrow{{}}\underset{1.12\,\,g}{\mathop{CaC{{l}_{2}}}}\,\] \[\left[ \begin{align} & 56\,\,\to \,\,100 \\ & 0.56\,\,\,\,\,\,1\,g \\ \end{align} \right]\] \[\left[ \begin{align} & 100\,\,\to \,\,112 \\ & 1\,g\,\,\,\,\,\,1.12 \\ \end{align} \right]\] \[\therefore \] Amount of \[NaCl=4.44-1.12=3.32\] \[%\] of \[NaCl=\frac{3.32}{4.44}\times 100=75\]You need to login to perform this action.
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