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3 years ago

How much liquid is needed to prepare 629.1mL of a solution that has a new concentration of 11.2M if the stock solution is 26.1M

Chemistry

1 answer:

slega [8]3 years ago

5 0

Answer:

270. mL

General Formulas and Concepts:

Acid-Base Titrations

Dilution: M₁V₁ = M₂V₂

M₁ is stock molarity
V₁ is stock volume
M₂ is new molarity
V₂ is new volume

Explanation:

Step 1: Define

Identify

[Given] V₂ = 629.1 mL

[Given] M₂ = 11.2 M

[Given] M₁ = 26.1 M

[Solve] V₁

Step 2: Find Stock Volume

Substitute in variables [Dilution]: (26.1 M)V₁ = (11.2 M)(629.1 mL)
Multiply: (26.1 M)V₁ = 7045.92 M · mL
Isolate V₁ [Cancel out units]: V₁ = 269.959 mL

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs as our lowest.

269.959 mL ≈ 270. mL

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From this we conclude that the element (X) is lanthanum (La) that has molecular weight, 138.667 g/mol.

Hence, the chemical symbol for this element is, (La)

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