Question

how many mol of na₃po₄ are in 175.0 g of na₃po₄?
○ 1.07
○ 0.937
○ 170.
○ 0.587

Explanation:

Step1: Calculate molar mass of $\ce{Na3PO4}$

Molar mass of $\ce{Na}$ = 22.99 g/mol, $\ce{P}$ = 30.97 g/mol, $\ce{O}$ = 16.00 g/mol.
Molar mass = $3\times22.99 + 30.97 + 4\times16.00$ = $68.97 + 30.97 + 64.00$ = 163.94 g/mol.

Step2: Use moles formula ($n = \frac{m}{M}$)

Given $m = 175.0$ g, $M = 163.94$ g/mol.
$n = \frac{175.0}{163.94} \approx 1.07$ mol.

Answer:

1.07 (corresponding to the option "1.07")