Question

Number of neutrons present in 9mg of o18 is

Answer 1

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

Answer : The number of neutrons in 9 mg of $^{18}O$ is $30.11\times 10^{20}$ neutrons.

Solution : Given,

Mass of $^{18}O$ = 9 mg = 0.009 g       (1 g = 1000 mg)

Molar mass of $^{18}O$ = 18 g/mole

First we have to calculate the moles of $^{18}O$.

$\text{ Moles of }^{18}O=\frac{\text{ Given mass of }^{18}O}{\text{ Molar mass of }^{18}O}=\frac{0.009g}{18g/mole}=0.0005mole=5\times 10^{-4}moles$

1 mole of $^{18}O$ has $6.022\times 10^{23}$ molecule of $^{18}O$

$5\times 10^{-4}$ moles of $^{18}O$ has $(6.022\times 10^{23})\times (5\times 10^{-4})=30.11\times 10^{19}$ molecule of $^{18}O$

Now we have to calculate the number of neutrons in $^{18}O$.

The number of neutrons present in 1 molecule of $^{18}O$ is,

Number of neutrons = Atomic mass - Atomic number = 18 - 8 = 10 neutrons

Number of neutrons present in 1 molecule of $^{18}O$ = 10 neutrons

Number of neutrons present in $30.11\times 10^{19}$ molecule of $^{18}O$ = $30.11\times 10^{20}$ neutrons

Therefore, the number of neutrons in 9 mg of $^{18}O$ is $30.11\times 10^{20}$ neutrons.