Question

A 10 mL solution of NaCl was treated with excess AgNO₃ to precipitate 0.4368g of AgCl. What was the molarity of NaCl?

Answer 1

The molarity of NaCl in the 10 mL solution is approximately 0.305 M, determined from the 0.4368 g of AgCl precipitated, due to their 1:1 ratio.

Step-by-step explanation:

To find the molarity of NaCl, you have to first calculated the moles of AgCl precipitated by dividing the given mass (0.4368 g) by the molar mass of AgCl (143.32 g/mol). Since AgCl and NaCl have a 1:1 ratio in terms of chloride ions (Cl-), the moles of AgCl are equivalent to the moles of NaCl in the original solution. To determine the molarity, we divided the moles of NaCl by the volume of the solution in liters (0.01 L). Thus, the molarity of NaCl in the 10 mL solution is approximately 0.305 M, reflecting its concentration.