From 48 gram of helium ,6×10^23atom are removed then find the mass of carbon atom which contain same number helium atom are left
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Answered by
7
Answer:
We know by Avogadro number that, 6.023 x 10^(23) no. of atoms of Helium make 1 mole that is equivalent to molar mass of Helium approximately 4 grams.
Thus, 6.023 x 10^(24) atoms of Helium = (1/(6.023 x 10^(23) ) x 6.023 x 10^(24) moles = 10 moles.
Since 1 mole = 4 grams(approx.) we can easily say 10 moles of helium will have a mass of 40 grams(approx.).
Explanation:
Answered by
12
Answer:
Initial no of moles=Given mass of He ÷ GAM
= 48 ÷ 4 = 12
(consider 6×10^23 is equal to 1 mole)
No of moles remaining =12-1 =11 moles
Given that
No of moles remaining (He) = No of carbon
11 = Wc/12
Wc= 11×12 =132 gm
hope it helps u
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