General Knowledge - Chemistry - Discussion

Discussion Forum : Chemistry - Section 1 (Q.No. 32)
32.
The mass of one Avogadro number of helium atom is
1.00 gram
4.00 gram
8.00 gram
4 x 6.02 x 1023 gram
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

Tenzin said:   1 decade ago
As we know that one mole is equal to 6.022 avogadro number or 1mole = 6.022amu so is the wt in gram. All three are related as one mole = gram wt of molecule = avogadro number that why He atom atomic number two and wt 4 so equal to 1amu.

Vijay said:   1 decade ago
The mass number of helium atom = 4.

As the avogadro number = 6.022*10^23.

Thus the mass of one avogadro number of helium atom = (4*6.022*10^23)/(6.022*10^23) = 4gm.
(1)

Versha said:   4 years ago
In question, Helium atom is given which means its elemental state (He) not its molecular state (He2). So, the correct answer of mass of helium atom is 2 gm not 4 gm.

Versha said:   4 years ago
In question, Helium atom is given which means its elemental state (He) not its molecular state (He2). So, the correct answer of mass of helium atom is 2 gm not 4 gm.

Bikash Ranjan Parhi said:   1 decade ago
Mass no of Helium atom is 4 gm.

One Avogadro no means 6.023*10^23 no of atoms = 4 gm of Helium.

So the mass of one Avogadro number of helium atom is 4.

Yusuf said:   9 years ago
One mole of helium is equal to one mole of avogardoes constant which is equal to 4 grams of helium.

Skp said:   6 years ago
That the Avogadrods no is divided by the mass no * Avogadro's number. Then answer is 4.00gm.

Preeti said:   1 decade ago
Mass of one mole that is Avogadro no. Is equal to the molecular mass in grams.

Rojalin said:   8 years ago
I too have the same doubt @Mira.

Please anyone clear our doubts.
(1)

Anom said:   6 years ago
It's said "1 Avogadro's number"! so 4.00 will be the answer.


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