Answer:
124225.91 g of Na₃PO₄
Explanation:
From the question given above, the following data were obtained:
Number of atoms of Na₃PO₄ = 4.56×10²⁶ atoms
Mass of Na₃PO₄ =?
From Avogadro's hypothesis,
6.02×10²³ atoms = 1 mole of Na₃PO₄
Next, we shall determine the mass of 1 mole of Na₃PO₄. This can be obtained as follow:
1 mole of Na₃PO₄ = (23×3) + 31 + (16×4)
= 69 + 31 + 64
= 164 g
Thus,
6.02×10²³ atoms = 164 g of Na₃PO₄
Finally, we shall determine the mass of Na₃PO₄ that contains 4.56×10²⁶ atoms. This can be obtained as follow:
6.02×10²³ atoms = 164 g of Na₃PO₄
Therefore,
4.56×10²⁶ atoms = (4.56×10²⁶ × 164)/6.02×10²³
4.56×10²⁶ atoms = 124225.91 g of Na₃PO₄
Therefore, 124225.91 g of Na₃PO₄ contains 4.56×10²⁶ atoms
