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Question Number 28076 by Tinkutara last updated on 20/Jan/18

Commented by Tinkutara last updated on 20/Jan/18

Answer given is 3rd. Is the answer correct?

Answered by ajfour last updated on 20/Jan/18

Atomic wt. = 2g/mol  no. of atoms =((4g)/(2g/mol))=2mol=2N_A   no. of neutrons =(1/(1atom))×2N_A atoms       =2N_A  .

$${Atomic}\:{wt}.\:=\:\mathrm{2}{g}/{mol} \\ $$$${no}.\:{of}\:{atoms}\:=\frac{\mathrm{4}{g}}{\mathrm{2}{g}/{mol}}=\mathrm{2}{mol}=\mathrm{2}{N}_{{A}} \\ $$$${no}.\:{of}\:{neutrons}\:=\frac{\mathrm{1}}{\mathrm{1}{atom}}×\mathrm{2}{N}_{{A}} {atoms} \\ $$$$\:\:\:\:\:=\mathrm{2}{N}_{{A}} \:. \\ $$

Commented by Tinkutara last updated on 20/Jan/18

Yes this is the same I got thanks but I wanted to confirm the answer.

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