The number of moles of Mg_3(PO_4)_2 containin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The chemical formula $Mg_3(PO_4)_2$ contains $8$ oxygen atoms per formula unit.Therefore,$1$ mole of $Mg_3(PO_4)_2$ contains $8$ moles of oxygen a

Vedclass · Accepted Answer

$3.125 	imes 10^{-2}$

Vedclass · Answer

(B) The chemical formula $Mg_3(PO_4)_2$ contains $8$ oxygen atoms per formula unit.Therefore,$1$ mole of $Mg_3(PO_4)_2$ contains $8$ moles of oxygen atoms.To find the moles of $Mg_3(PO_4)_2$ containing $0.25$ moles of oxygen atoms,we use the ratio:$	ext{Moles of } Mg_3(PO_4)_2 = \frac{	ext{Moles of oxygen atoms}}{8}$$	ext{Moles of } Mg_3(PO_4)_2 = \frac{0.25}{8} = 0.03125 = 3.125 	imes 10^{-2}$.

The number of moles of $Mg_3(PO_4)_2$ containing $0.25$ moles of oxygen atoms is:

Similar Questions

Calculate the number of molecules in a gas sample having a volume of $1.12 \times 10^{-7} \, cc$ at $NTP$.

Which of the following contains the maximum number of atoms?

$8 \ g$ of $O_2$ has the same number of atoms as that in:

If $10^{21}$ molecules are removed from $200 \, mg$ of $CO_2$,then the number of moles of $CO_2$ left is .....

The mass of a molecule of water is

Vedclass Test Series

Exam Paper Generator

Online Exam Module