Answer:
86 mL
Explanation:
First find the moles of Pb (NO3)2
n=cv
where
c ( concentration)= 0.210 M
v ( volume in L) =0.05
n= 0.210 × 0.05
n= 0.0105
Using the mole ratio, we can find the moles of KCl by multiplying by 2
n (KCl) =0.0105 ×2
=0.021
v (KCl)= n/ c
= 0.021/ 0.244
=0.08606557377
=0.086 L
= 86 mL
Answer:
The volume of 0.244 M KCl solution is required 86.07 mL.
Explanation:
[tex]Concentration=\frac{Moles}{Volume (L)}[/tex]
Moles of lead nitrate in 50.0 mL of 0.210 M solution be n.
Molarity of the lead nitrate solution = 0.210 M
Volume of the solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)
[tex]0.210 M=\frac{n}{0.050 L}[/tex]
[tex]n=0.210 M\times 0.050 L=0.0105 mol[/tex]
According to reaction given, 1 mole of lead nitrate reacts with 2 moles of KCl.
Then 0.0105 moles of lead nitrate will react with:
[tex]\frac{2}{1}\times 0.0105 mol=0.021 mol[/tex] of KCl
Moles of KCl = 0.021 mol
Volume of KCl solution = V
Molarity of the KCl solution = 0.244 M
[tex]0.244 M=\frac{0.021 mol}{V}[/tex]
[tex]V=\frac{0.021 mol}{0.244 M}=0.08607 L = 86.07 mL[/tex]
The volume of 0.244 M KCl solution is required 86.07 mL.
