Question

In an experiment, a 88.11 mL sample of unknown silver nitrate solution was treated with 9.753 g of sodium chloride, resulting in 4.576 g of precipitate. Calculate the molarity of the silver nitrate solution

Answer 1

Answer:

0.362 M

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 * $\frac{1molAgNO_3}{1molAgCl}$ = 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