Respuesta :
Answer:
0.1891 L minimum volume of 0.200 M potassium iodide solution is required.
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Or,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
Given :
For lead(II) nitrate solution :
Molarity = 0.122 M
Volume = 155.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 155.0×10⁻³ L
Thus, moles of lead(II) nitrate :
[tex]Moles=0.122 \times {155.0\times 10^{-3}}\ moles[/tex]
Moles of lead(II) nitrate = 0.01891 moles
According to the given reaction:
[tex]2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow 2KNO_3_{(aq)}+2PbI_2_{(aq)}[/tex]
1 mole of lead(II) nitrate reacts with 2 moles of potassium iodide
0.01891 mole of lead(II) nitrate reacts with 2*0.01891 moles of potassium iodide
Moles of potassium iodide = 0.03782 moles
Also,
Molarity = 0.200 M
Volume = ?
[tex]Volume=\frac{Moles\ of\ solute}{Molarity}[/tex]
[tex]Volume=\frac{0.03782}{0.200}\ L=0.1891\ L[/tex]
0.1891 L minimum volume of 0.200 M potassium iodide solution is required.
