Respuesta :
Answer: The volume of silver nitrate solution required is 0.474 L
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
- For sodium chloride:
Molarity of sodium chloride = 0.12 M
Volume of solution = 0.10 L
Putting values in above equation, we get:
[tex]0.12M=\frac{\text{Moles of sodium chloride}}{0.10L}\\\\\text{Moles of sodium chloride}=(0.12mol/L\times 0.10L)=0.012mol[/tex]
1 mole of sodium chloride produces 1 mole of chloride ions and 1 mole of sodium ions
So, moles of chloride ions = 0.012 moles
- For magnesium chloride:
Molarity of magnesium chloride = 0.18 M
Volume of solution = 0.23 L
Putting values in above equation, we get:
[tex]0.18M=\frac{\text{Moles of magnesium chloride}}{0.23L}\\\\\text{Moles of magnesium chloride}=(0.18mol/L\times 0.23L)=0.0414mol[/tex]
1 mole of magnesium chloride produces 1 mole of magnesium ions and 2 moles of chloride ions
So, moles of chloride ions = [tex](2\times 0.0414)=0.0828moles[/tex]
Total moles of chloride ions = [0.012 + 0.0828] = 0.0948 moles
The chemical equation for the reaction of chloride ions and silver nitrate follows:
[tex]AgNO_3(aq.)+Cl^-(aq.)\rightarrow AgCl(s)+NO_3^-(aq.)[/tex]
By Stoichiometry of the reaction:
1 mole of chloride ions reacts with 1 mole of silver nitrate
So, 0.0948 moles of chloride ions will react with = [tex]\frac{1}{1}\times 0.0948=0.0948mol[/tex] of silver nitrate
- For silver nitrate:
Molarity of silver nitrate = 0.200 M
Moles of silver nitrate = 0.0948 moles
Putting values in above equation, we get:
[tex]0.200M=\frac{0.0948mol}{\text{Volume of silver nitrate}}\\\\\text{Volume of silver nitrate}=\frac{0.0948mol}{0.200}=0.474L[/tex]
Hence, the volume of silver nitrate solution required is 0.474 L
The volume of silver nitrate solution required is 0.474 L
The molarity, number of moles and volume of a solution are related by the formula:
number of moles = molarity * volume
The moles of chloride ions in each solution is first determined:
number of moles of NaCl = 0.12 * 0.1 = 0.012 moles of NaCl
number of moles of MgCl₂ = 0.18 * 0.23 = 0.0414 moles of MgCl₂
From the dissociation equations of sodium chloride and magnesium chloride:
NaCl (aq) ----> Na⁺(aq) + Cl⁻(aq)
MgCl₂ (aq) ----> Mg²⁺(aq) + 2 Cl⁻(aq)
1 mole of NaCl produces 1 mole of Cl⁻
0.012 moles of NaCl will produce 0.012 moles of Cl⁻
1 mole of MgCl₂ produces 2 moles of Cl⁻
0.0414 moles of MgCl₂ will produce 2 * 0.0414 moles of Cl⁻ = 0.0828 moles of Cl⁻
Total moles of Cl⁻ ions = 0.012 + 0.0828 = 0.0948 moles of Cl⁻ ions
The volume of silver nitrate solution required is then calculated:
The equation of the reaction between silver ion and chloride ion is given below:
Ag⁺ (aq) + Cl⁻ (aq) ----> AgCl (s)
1 mole of chloride ion reacts with one of silver ion
0.0948 moles of chloride ion will react with 0.0948 moles of silver ions.
AgNO₃ (aq) ----> Ag⁺ (aq) + NO₃⁻ (aq)
1 mole of AgNO₃ produces 1 mole of Ag⁺
0.0948 moles of AgNO₃ will produce 0.0948 moles of Ag⁺.
Volume of silver nitrate solution required is then obtained from the formula:
Volume = number of moles / molarity
Volume = 0.0948 /0.200 M = 0.474 L
Therefore, the volume of silver nitrate solution required is 0.474 L
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