question_answer

The number of \[\pi \] electrons present in 6.4 g of calcium carbide is - (\[{{N}_{A}}=\] Avagadro's number)

A)  \[4\,{{N}_{A}}\]                      

B)  \[0.4\,{{N}_{A}}\]

C)  \[0.1\,{{N}_{A}}\]                   

D)  \[0.2\,{{N}_{A}}\]

Correct Answer: B

Solution :

 \[6.4g\] of \[Ca{{C}_{2}}\] contain \[\pi \]-electron \[=4{{N}_{2}}\]