A titration process is done by using 0.2M of 20 ml of third analytical group main reagent solution with 0.1 M nitric acid soluti
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Answer:
The concentration of NaCl = 0.3374 M
Explanation:
Given :
Molarity of AgNO₃ = 0.2503 M
Volume of AgNO₃ = 20.22 mL
The conversion of mL into L is shown below:
Thus, volume of the solution = 20.22×10⁻³ L
Molarity of a solution is the number of moles of solute present in 1 L of the solution.
The formula can be written for the calculation of moles as:
Thus,
The chemical reaction taking place:
According to reaction stoichiometry:
<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl
Thus,
5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl
Thus, moles of NaCl required = 5.0611×10⁻³ moles
Volume of NaCl required = 15.00 mL
The conversion of mL into L is shown below:
Thus, volume of the solution = 15.00×10⁻³ L
Applying in the formula of molarity as:
<u>Thus, the concentration of NaCl = 0.3374 M</u>