Question

No. Of oxygen atoms present in 14.6 g of magnesium bicarbonate

Answer 1

Hence 0.51Na oxygen atoms will be there

Answer 2

Answer : The number of oxygen atoms present in 14.6 g of magnesium bicarbonate is $3.6132\times 10^{23}$.

Solution : Given,

Mass of magnesium bicarbonate = 14.6 g

Molar mass of magnesium bicarbonate = 146 g/mole

The chemical formula of magnesium bicarbonate is, $Mg(HCO_3)_2$.

First we have to calculate the moles of magnesium bicarbonate.

Formula used :  $Moles=\frac{\text{ Given mass}}{\text{ Molar mass}}$

Moles of $Mg(HCO_3)_2$ = $\frac{14.6g}{146g/mole}=0.1moles$

Number of oxygen atom present in $Mg(HCO_3)_2$ = 6

In 1 mole of $Mg(HCO_3)_2$, number of atoms present = $(6)\times (6.022\times 10^{23})$

In 0.1 mole of $Mg(HCO_3)_2$, number of atoms present = $(6)\times (6.022\times 10^{23})\times 0.1=3.6132\times 10^{23}$

Therefore, The number of oxygen atoms present in 14.6 g of magnesium bicarbonate is $3.6132\times 10^{23}$.