Question

14 gi of ( C_{14} ) cosstain ( 6.023 times 10^{23} ) no. of ( C_{14} ) atoms. ( L 4 ) g of ( c_{14} ) contain ( 6,023 times 10^{23} times 8 ) no. of neutrons
1 g of Ciq comtein ( frac{6,023 times 10^{23} times 8}{14} ) no. of heutrous
So ( 7 m gleft(i cdot e cdot frac{7}{1000} gright) C_{14} ) contain ( frac{6 cdot 023 times 10^{23} times 8^{4} times 71}{142} times 1000 )
[
begin{array}{l}
=6.023 times 4 times 10^{23-3}
=24 cdot 0.92 times 10^{20}
=2.409 times 10^{21} mathrm{no.} text { of } A text { extrom }
end{array}
]

# no of nutrones in Tong 0), Calculate of the Cu

Solution