A hypothetical element 'Coronium' has two isotopes $$A_1$$ and $$A_2$$. The mass of $$3 \times 10^{22}$$ atoms of $$A_1$$ is 20 g while the mass of $$1.5 \times 10^{22}$$ atoms of $$A_2$$ is 10.5 g. Then average atomic weight of element 'Coronium' in isotopic mixture ($$3 \times 10^{22}$$ atoms of $$A_1$$ and $$1.5 \times 10^{22}$$ atoms of $$A_2$$) would be approximately

A

418.30

B

408.23

C

406.67

D

data insufficient