Question

If 25.8 ml of agno3 solution is needed to precipitate all the cl- ions in a 785 mg sample of kcl (forming agcl) what is the molarity of the agno3 solution

Answer 1

Answer : The molarity of the AgNO₃ solution is 0.408 M

Answer :

The balanced chemical equation for the reaction between KCl and AgNO₃ is given below.

$KCl + AgNO_{3} \rightarrow AgCl + KNO_{3}$

Step 1 : Find moles of KCl.

The mass of KCl is given as 785 mg. Let's convert this to grams.

$785 mg \times \frac{1g}{1000 mg} = 0.785 g$

The molar mass of KCl is 74.55 g/mol

The moles of KCl can be calculated as,

$mol KCl = \frac{grams}{MolarMass}$

$mol KCl = \frac{0.785g}{74.55 g/mol} = 0.0105$

We have 0.0105 mol KCl

Step 2 : Find moles of AgNO₃.

The mole ratio of KCl and AgNO₃ is 1 : 1.

The moles of AgNO₃ can be calculated using this mole ratio.

$mol AgNO_{3} = 0.0105 mol KCl \times \frac{1 mol AgNO_{3}}{1 mol KCl} = 0.0105$

We have 0.0105 mol AgNO₃.

Step 3 : Use molarity formula.

The molarity of AgNO₃ can be calculated as,

$Molarity = \frac{mol AgNO_{3}}{L }$

The volume of AgNO₃ solution is 25.8 mL which is 0.0258 L.

$Molarity = \frac{0.0105mol}{0.0258L} = 0.408 M$

The molarity of the AgNO₃ solution is 0.408 M.