calculate no. of atom of constituent elements in 53 g of Na2CO3
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Answered by
10
HEY mate here is your answer.
hope it helps you
hope it helps you
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Anonymous:
so it should be multiplied by the atomicity
hmm
kk
0.25*avagadro number
got it
yes but number of moles is wrong
Yeah it was bt now I have edited that
siril
ur is wrong siril i think so
so be it...whatever you say
Answered by
8
number of moles in 53 grams of Na2CO3 = 53/106 = 0.5
Now, 1molecule of Na2CO3 contains 6 atoms (2 Na, 1 C, 3 O)
wherein 1 mole of Na2CO3 contains "6 x Avogadro number" of atoms
Therefore 0.5moles of Na2CO3 contains "0.5(6 x Avogadro number)" of atoms
= 0.5 (6 x 6 x 10²³) atoms
= 0.5 (36 x 10²³) atoms
= 18 x 10²³ atoms
= 1.8 x 10²⁴ atoms
Now, 1molecule of Na2CO3 contains 6 atoms (2 Na, 1 C, 3 O)
wherein 1 mole of Na2CO3 contains "6 x Avogadro number" of atoms
Therefore 0.5moles of Na2CO3 contains "0.5(6 x Avogadro number)" of atoms
= 0.5 (6 x 6 x 10²³) atoms
= 0.5 (36 x 10²³) atoms
= 18 x 10²³ atoms
= 1.8 x 10²⁴ atoms
your answer is correct if the question was "find the number of molecules of Na2CO3"
hm
you must multiply it with 6 to get the correct answer for the number of atoms
no
acc. to formula
its of atoms
siril is correct
to calculate the number of atoms you must multiply ur answer with 6
dev you have to multiply mole with the atomicity
because each molecule of Na2CO3 contains 6 atoms...
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