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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)

Suppose A Lab Procedure Calls For 075 L Of A 025 M CaCl2 Solution How Much Of A 100 M Stock Solution Do We Dilute Where The Diluted Solution Is 2 The Numerator class=

Answer :

Neetoogo

Answer:

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

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