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How many ml of 10.0 m cacl2 do we need to prepare 3.00 l of a 0.500 m cacl2 solution

User Kamalpreet
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1 Answer

4 votes

Answer:

150 mL

Step-by-step explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 10 M

Volume of diluted solution (V₂) = 3 L

Molarity of diluted solution (M₂) = 0.5 M

Volume of stock solution needed (V₁) =?

The volume of stock solution needed can be obtained as follow:

M₁V₁ = M₂V₂

10 × V₁ = 0.5 × 3

10 × V₁ = 1.5

Divide both side by 10

V₁ = 1.5 / 10

V₁ = 0.15 L

Finally, we shall convert 0.15 L to mL. This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.15 L = 0.15 L × 1000 mL / 1 L

0.15 L = 150 mL

Therefore, the volume of the stock solution needed is 150 mL

User Chara
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