Question

calculate Number of electron present in 9.5g of po43-

Answer 1

Moles Of P O_4^{-3} [/tex] = Mass in g / molar mass of component 
                                           = $\frac{9.5}{31+4×16 }$

Answer 2

Answer: Number of electrons present is $301.1\quad \times \quad { 10 }^{ 22 }\quad atoms$

From the given,

Mass of ${ PO }_{ 4 }^{ 3- }$ = 9.5 g

Let's calculate number of moles of ${ PO }_{ 4 }^{ 3- }$

             $Moles\quad of\quad { PO }_{ 4 }^{ 3- }\quad =\quad \frac { Mass\quad in\quad grams }{ Molar\quad mass\quad of\quad { PO }_{ 4 }^{ 3- } }$

              $=\quad \frac { 9.5 }{ 31\quad +\quad 4\quad \times \quad 16 }$

              $Moles\quad of\quad { PO }_{ 4 }^{ 3- }\quad =\quad 0.1\quad mole$

              $1\quad mole\quad contains\quad 6.022\quad \times\quad { 10 }^{ 23}\quad atoms$

              $0.1\quad mole\quad of\quad { PO }_{ 4 }^{ 3- }\quad contains\quad =\quad 0.1\quad \times \quad 6.022\quad \times { 10 }^{ 23 }\quad atoms\quad \\=\quad 6.022\quad \times { 10 }^{ 22 }\quad atoms$

              $1\quad molecule\quad of\quad { PO }_{ 4 }^{ 3- }\quad contains\quad 15\quad +\quad 4\quad \times \quad 8\quad +3\quad =\quad 50\quad electrons$

              $=\quad 6.022\quad \times \quad { 10 }^{ 22 }\quad atoms\quad \times \quad 50$

              $=\quad 301.1\quad \times \quad { 10 }^{ 22 }\quad atoms$