General Knowledge - Chemistry - Discussion
Discussion Forum : Chemistry - Section 1 (Q.No. 32)
32.
The mass of one Avogadro number of helium atom is
Discussion:
16 comments Page 2 of 2.
Mira said:
10 years ago
Why should we divide 6.023*10^23?
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.
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.
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.
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)
Preeti said:
1 decade ago
Mass of one mole that is Avogadro no. Is equal to the molecular mass in grams.
RAVI said:
1 decade ago
Always mass is just double of atomic number.
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.
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