Answered

What is the molar mass of [tex]$\text{Mg}_3\left(\text{PO}_4\right)_2$[/tex], a substance formerly used in medicine as an antacid? Use the given atomic masses in the Periodic Table (do not round those values) and round your final answer to one decimal place.

A. [tex]214.3 \, \text{g/mol}[/tex]
B. [tex]262.9 \, \text{g/mol}[/tex]
C. [tex]150.3 \, \text{g/mol}[/tex]
D. [tex]118.3 \, \text{g/mol}[/tex]
E. [tex]71.3 \, \text{g/mol}[/tex]



Answer :

To determine the molar mass of the compound [tex]\( \text{Mg}_3(\text{PO}_4)_2 \)[/tex], we need to consider the atomic masses of its constituent elements and calculate the total molar mass of the compound. Here is a step-by-step solution:

1. Identify the formula and its components: [tex]\( \text{Mg}_3(\text{PO}_4)_2 \)[/tex]:
- This formula indicates that there are 3 magnesium (Mg) atoms, 2 phosphorus (P) atoms, and 8 oxygen (O) atoms in one formula unit of the compound.

2. Determine the atomic masses of each element (as given in the problem statement):
- Atomic mass of magnesium (Mg): 24.305 g/mol
- Atomic mass of phosphorus (P): 30.974 g/mol
- Atomic mass of oxygen (O): 15.999 g/mol

3. Calculate the contribution of each type of atom to the molar mass:
- For magnesium: There are 3 Mg atoms.
[tex]\[ \text{Mass contribution of Mg} = 3 \times 24.305 \, \text{g/mol} = 72.915 \, \text{g/mol} \][/tex]
- For phosphorus: There are 2 P atoms.
[tex]\[ \text{Mass contribution of P} = 2 \times 30.974 \, \text{g/mol} = 61.948 \, \text{g/mol} \][/tex]
- For oxygen: There are 8 O atoms.
[tex]\[ \text{Mass contribution of O} = 8 \times 15.999 \, \text{g/mol} = 127.992 \, \text{g/mol} \][/tex]

4. Sum the contributions to find the total molar mass of the compound:
[tex]\[ \text{Molar mass of } \text{Mg}_3(\text{PO}_4)_2 = 72.915 \, \text{g/mol} + 61.948 \, \text{g/mol} + 127.992 \, \text{g/mol} = 262.855 \, \text{g/mol} \][/tex]

5. Round the final molar mass to one decimal place:
[tex]\[ \text{Molar mass of } \text{Mg}_3(\text{PO}_4)_2 = 262.9 \, \text{g/mol} \][/tex]

The molar mass of [tex]\( \text{Mg}_3(\text{PO}_4)_2 \)[/tex] rounded to one decimal place is [tex]\( 262.9 \, \text{g/mol} \)[/tex].

Hence, the correct answer is [tex]\( 262.9 \, \text{g/mol} \)[/tex].