The number of electron in 3.1 mg NO3- is
(1) 32
(2) 1.6 x 10-
(3) 9.6 x 1020
(4) 9.6 x 1023
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Answered by
0
Answer:
it's option ( 2) 1.6 x 10^ 20.
Explanation:
atomicity of NO3- = 4
NA = 6.023 X 10 ^ 23
no.of moles = gram / molecular mass = 0.0031 / 64 * where ( 1g = 1000mg)
Total no.of e- = moles x NA X atomicity.
= 0.0031 / 64 x 6 x 10 ^23 x 4
= 1.6 x 10 ^ 20.
Answered by
0
Answer:
correct answer is 9.6×10^-5 .
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