Question

no.of neutrons present in 9mg of o18

Answer 1

here is your ans...

Here no of neutron in one atom of 18O=18-8=10.Therefore no of neutron in 18 g of 18O=10×Na.So no of neutron in 9mg of O18=(10×Na×9×10^-3)÷18=3×10^21 So answer is 3×10^21..


hope it helps..

Answer 2

The molar mass of O¹⁸ is 18grams

The sample of O¹⁸ taken weighs 9milligrams = 0.009 grams

so,  number of moles present in the sample = 0.0005 moles

1 atom of O¹⁸ contains 10 neutrons
where 1mole contains Avogadro number(N₀) of atoms

Thus, 1mole contains '10x(N₀)' of neutrons

Hence 0.0005 moles contains '0.0005 x 10N₀' number of neutrons

Therefore the given sample contains 0.005N₀ number of neutrons