Question

calculate number of electron present in 9.5g of po4'3-

Answer 1

so u can find this formula...

electron==mol×6.022×10^23×number of electron in po4-3

Answer 2

Answer : The number of electrons in 9.5 g of $PO^{3-}_4$ is $240\times 10^{22}$ electrons.

Solution : Given,

Mass of $PO^{3-}_4$ = 9.5 g

Molar mass of $PO_4$ = 95 g/mole

First we have to calculate the moles of $PO^{3-}_4$.

Moles of $PO^{3-}_4$ = $\frac{\text{ Given mass}}{\text{ Molar mass}}= \frac{9.5g}{95g/mole}=0.1mole$

1 mole of $PO^{3-}_4$ has $6.022\times 10^{23}$ molecule of $PO^{3-}_4$

0.1 mole of $PO^{3-}_4$ has $(6.022\times 10^{23})\times (0.1)=6.022\times 10^{22}$ molecule of $PO^{3-}_4$

Now we have to calculate the number of electrons in $PO^{3-}_4$.

The number of electrons present in 1 molecule of $PO^{3-}_4$ is,

= (5) + 4(8) + 3 = 40 electrons

Number of electrons present in 1 molecule of $PO^{3-}_4$ = 40 electrons

Number of electrons present in $6.022\times 10^{22}$ molecule of $PO^{3-}_4$ = $(6.022\times 10^{22})\times (40)=240\times 10^{22}$ electrons

Therefore, the number of electrons in 9.5 g of $PO^{3-}_4$ is $240\times 10^{22}$ electrons.