Answer:Hope this helps
Explanation:
t is given by the formula for dilution or neutralization reaction:
n₁*M₁*V₁= n₂*M₂*V₂
where,
n₁, M₁ ,V₁ are the n-factor, molarity and volume of acid
n₂M₂ ,V₂ are the n-factor, molarity and volume of base
For a:
Given:
n₁=1
M₁=0.105M
n₂ =1
M₂=0.00950M
V₂= 50.0mL
To find: V₁=?
On substituting the values:
n₁*M₁*V₁= n₂*M₂*V₂
1*0.105*V₁=1*0.00905*50
V₁=4.52mL
Hence, the volume of HCl needed is 4.52 mL
For b:
Given:
n₁=1
M₁=0.105M
n₂ =1
M₂=0.117 M
V₂= 23.0 mL
To find: V₁=?
On substituting the values:
n₁*M₁*V₁= n₂*M₂*V₂
1*0.105*V₁=1*0.117 * 23
V₁=25.63 mL
Hence, the volume of HCl needed is 25.63 mL
For c:
Given:
Mass of NaOH = 1.40 g
Molar mass of NaOH = 40 g/mol
Volume of solution = 1 L
On substituting the values:
Molarity of NaOH = 1.40 / 40
Molarity of NaOH = 0.035M
Given:
n₁=1
M₁=0.105M
n₂ =1
M₂=0.035 M
V₂= 130 mL
To find: V₁=?
On substituting the values:
n₁*M₁*V₁= n₂*M₂*V₂
1*0.105*V₁=1*0.035 * 130
V₁=43.33 mL
Hence, the volume of HCl needed is 43.33 mL.
