Question

the number of electron in 3.1 mg No3 ^-is

Answer 1

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Answer 2

Answer : The number of electrons in 3.1 mg moles of $NO^-_3$ is $9.64\times 10^{20}$electrons.

Solution : Given,

Mass of $NO^-_3$ = 3.1 mg

Molar mass of $NO_3$ = 62 g/mole

First we have to calculate the moles of $NO^-_3$.

Moles of $NO^-_3$ = $\frac{\text{ Given mass}}{\text{ Molar mass}}= \frac{3.1\times 10^{-3}g}{62g/mole}=0.05\times 10^{-3}moles$

1 mole of $NO^-_3$ has $6.022\times 10^{23}$ molecule of $NO^-_3$

0.05 moles of $NO^-_3$ has $(6.022\times 10^{23})\times 0.05=0.3011\times 10^{20}$ molecule of $NO^-_3$

Now we have to calculate the number of electrons in $NO^-_3$.

The number of electrons present in 1 molecule of $NO^-_3$ is,

$NO^-_3$ = (7) + 3(8) + 1 = 32 electrons

Number of electrons present in 1 molecule of $NO^-_3$ = 32 electrons

Number of electrons present in  molecule of $NO^-_3$ = $9.64\times 10^{20}$ electrons

Therefore, the number of electrons in 3.1 mg moles of $NO^-_3$ is $9.64\times 10^{20}$ electrons.