Question

If 37.1 mL AgNO3 solution reacts with excess potassium chloride solution to yield 1.56 g of AgCl precipitate, what is the molarity of silver ion in the original solution? AgCl = 143.3 g/mol Enter your answer in decimal format with three decimal places and no units.

Answer 1

The molarity of silver ion in the original AgNO3 solution is 0.293 M.

Step-by-step explanation:

To determine the molarity of silver ion in the original AgNO3 solution when 37.1 mL reacts to yield 1.56 g of AgCl, we first need to calculate the moles of AgCl produced using the molar mass of AgCl.

The molar mass of AgCl is 143.3 g/mol. So, the moles of AgCl formed are calculated as follows:

(1.56 g AgCl) / (143.3 g/mol) = 0.010878 mol AgCl

Since AgNO3 reacts with KCl in a 1:1 mole ratio to produce AgCl, the moles of AgNO3 that reacted is also 0.010878 mol. To find the molarity of AgNO3, we divide the moles by the volume of solution in liters:

(0.010878 mol AgNO3) / (0.0371 L) = 0.293 M

Answer 2

Molarity of silver ion: 0.296 M

Step-by-step explanation:

Reaction:

AgNO₃ + KCl → AgCl↓ + K⁺ + NO₃⁻

From the reaction, we know that the moles of AgCl produced will be the same as the moles of initial silver.

First, let´s calculate the number of moles of AgCl produced:

1.56 g AgCl was produced, that is, (1.56 g AgCl * 1 mol AgCl/143.3 g AgCl) 0.011 moles AgCl.

The moles of silver ion present in the original solution was 0.011 mol. Since this number of moles was present in a 37.1 ml solution, then, in 1000 ml:

moles of silver ion per liter = 1000 ml * 0.011 mol / 37.1 ml = 0.296 mol

Molarity of silver ion = 0.296 M