Answer:
840mL
Explanation:
Given parameters:
Concentration of stock = 5M
New volume = 3L
New concentration = 1.40M
Unknown:
Volume of stock needed = ?
Solution:
A concentrated standard solution which is the stock may become diluted to a lower concentration before using.
One important thing is that the number of moles of the substance in the solution before and after is the same.
Therefore;
C₁ V₁ = C₂ V₂
where C and V are concentration and volume
1 and 2 are initial and final states.
C₁ = concentration of stock
V₁ = volume of stock
C₂ = final concentration
V₂ = final volume
5 x V₁ = 1.4 x 3
V₁ = 0.84L
This gives 840mL
