Answer:
0.3714 M
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
AgNO3 (aq) + NaCl (aq) -> AgCl(s) +
NaNO3 (aq)
Step 2:
Determination of the number of moles of AgNO3 in 30.00 mL of 0.2503 M AgNO3. This is illustrated below:
Molarity of AgNO3 = 0.2503 M
Volume of solution = 30mL = 30/1000 = 0.03L
Number of mole of AgNO3 =?
Molarity = mole /Volume
0.2503 = mole / 0.03
Cross multiply to express in linear form
Mole = 0.2503 x 0.03
Mole of AgNO3 = 7.509x10^-3 mole
Step 3:
Determination of the number of mole of NaCl required to react with AgNO3. This is illustrated below:
AgNO3 (aq) + NaCl (aq) -> AgCl(s) +
NaNO3 (aq)
From the balanced equation above,
1 mole of AgNO3 required 1 mole of NaCl for complete neutralization.
Therefore, 7.509x10^-3 mole of AgNO3 will also require 7.509x10^-3 mole of NaCl for complete neutralization.
Step 4:
Determination of the molarity of NaCl. This is illustrated below:
Mole of NaCl = 7.509x10^-3 mole
Volume of solution = 20.22 mL = 20.22/1000 = 0.02022L
Molarity =?
Molarity = mole /Volume
Molarity of NaCl = 7.509x10^-3/0.02022
Molarity of NaCl = 0.3714 M
Answer:
The concentration of NaCl in the solution is 0.371 M
Explanation:
Step 1: Data given
Volume of 0.2503 M AgNO3 solution = 30.00 mL = 0.030 L
Volume of NaCl = 20.22 mL = 0.02022 L
Step 2: The balanced equation
AgNO3(aq) + NaCl(aq) ===> AgCl(s) + NaNO3(aq)
Step 3: Calculate concentration of NaCl
C1*V1 = C2*V2
⇒with C1 = the concentration of AgNO3 = 0.2503 M
⇒with V1= the volume of AgNO3 = 0.030 L
⇒with C2 = the concentration of NaCl = TO BE DETERMINED
⇒with V2 = the volume of NaCl = 0.02022 L
0.2503 M * 0.030 L = C2 * 0.02022L
C2 = (0.2503 * 0.030) / 0.02022
C2 = 0.371 M
The concentration of NaCl in the solution is 0.371 M
