Suppose a lab procedure calls for 0.75 L of a 0.25 M CaCl2 solution. How much of a 10.0 M stock solution do we dilute? Where the diluted solution is #2 (the numerator)

V₁ = 0.019 L

Explanation:

Given data:

Molarity of stock solution = 10.0 M

Volume of stock solution = ?

Molarity of CaCl₂ = 0.25 M

Volume of CaCl₂ = 0.75 L

Solution:

Formula:

V₁ = M₂V₂ / M₁

V₁ = 0.25 M × 0.75 L / 10.0 M

V₁ = 0.1875 M.L / 10.0 M

V₁ = 0.019 L

what are the óptions. pléase list them

it means there are 3 atoms of oxygen