A) 0.75
B) 0.5
C) 0.40
D) 0.25
Correct Answer: B
Solution :
[b] Let weight of C be xg, then S will be \[\left( 14-x \right)g\] \[\frac{x/12}{\left( 14-x \right)/32}=\frac{2}{1}\] \[\therefore x=6g\]; Moles of \[C=\frac{6}{12}=0.5\]You need to login to perform this action.
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