Answer:
0.362 M
Step-by-step explanation:
The reaction that takes place is:
- AgNO₃ + NaCl → NaNO₃ + AgCl(s)
First we convert the mass of AgCl (the precipitate) into moles, using its molar mass:
- 4.576 g AgCl ÷ 143.32 g/mol = 0.0319 mol AgCl
Now we convert AgCl moles into AgNO₃ moles:
- 0.0319 mol AgCl *
= 0.0319 mol AgNO₃
Finally we calculate the molarity, after converting 88.11 mL to L:
- 88.11 mL / 1000 = 0.08811 L
- Molarity = 0.0319 mol AgNO₃ / 0.08811 L = 0.362 M