A) 22kg
B) 18 kg
C) 9 kg
D) 27 kg
Correct Answer: B
Solution :
\[\frac{wt.\,of\,Ca}{wt.\,of\,Al}=\frac{Eq.\,wt.\,of\,Ca}{Eq.\,Wt.\,of\,Al}\] \[\frac{40}{wt.\,of\,Al}=\frac{{\scriptstyle{}^{40}/{}_{2}}}{{\scriptstyle{}^{27}/{}_{3}}}\] Wt. of \[Al=18\,kg\]
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