Number of atoms present in 52 g of helium is:

(N_A is Avogadro’s constant
Gram atomic mass of He is 4 g)

26 X N_A

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13 X N_A

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52 Χ N_A

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N_A

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Solution

The correct option is B
13 X N_A

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Similar questions

Number of atoms present in $52 g$ of helium is:
$(N_{A}$ is Avogadro’s constant, Gram atomic mass of helium is $4 g)$

N_{A}

The number of moles present in 52g of He(Atomic mass, He = 4) are:

181.60 mL

0^{\circ} C

1 bar

1 g

(N_{A}

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