The highest number of helium atoms is in

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(D) $(1)$ $4 \ u$ of $He = \frac{4 \ u}{4 \ u} = 1 \ He$ atom$(2)$ $4 \ g$ of helium $= \frac{4 \ g}{4 \ g \ mol^{-1}} = 1 \ mole = N_A \ He$ atoms$(3

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$4 \ mol$ of helium

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(D) $(1)$ $4 \ u$ of $He = \frac{4 \ u}{4 \ u} = 1 \ He$ atom$(2)$ $4 \ g$ of helium $= \frac{4 \ g}{4 \ g \ mol^{-1}} = 1 \ mole = N_A \ He$ atoms$(3)$ $2.271098 \ L$ of $He$ at $STP = \frac{2.271098}{22.71098} \ mol = 0.1 \ mol = 0.1 \ N_A \ He$ atoms$(4)$ $4 \ mol$ of $He = 4 \ N_A \ He$ atomsComparing the values,$4 \ mol$ contains the highest number of atoms.

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