Answer:
The required milliliters of CsNO3 in solution 1 = 42.29 mL
Explanation:
Given that:
The molarity of CsNO3 in solution 1 M₁= 0.266 M
The Volume of CsNO3 in solution 1 V₁ = ???
The Volume of CsNO3 in solution 2 V₂ = 150.0 mL
The molarity of CsNO3 in solution 2 M₂= 0.075 M
By the application of the titrimetric concept:
M₁V₁ = M₂V₂
We have;
0.266 M × V₁ = 0.075 M × 150.0 mL
V₁ = ( 0.075 M × 150.0 mL ) / 0.266 M
V₁ = (11.25 / 0.266 ) mL
V₁ = 42.29 mL
