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tamaranim1 [39]
1 year ago
11

Suppose you wish to make 0.879 l of 0.250 m silver nitrate by diluting a stock solution of 0.675 m silver nitrate. how many mill

iliters of the stock solution would you need to use?
Chemistry
1 answer:
sveticcg [70]1 year ago
3 0

Answer is: 325.5 mL of the stock solution.

c₁(AgNO₃) = 0.675 M.

<span> V</span>₂(AgNO₃) = 0.879 L.<span>
c</span>₂(AgNO₃) = 0.250 M.<span>
V</span>₁(AgNO₃) = ?<span>
c</span>₁ - original concentration of the solution, before it gets diluted.<span>
c</span>₂ - final concentration of the solution, after dilution.<span>
V</span>₁ - <span>volume to be diluted.
V</span>₂ - <span>final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂.<span>
V</span>₁(AgNO₃) = c₂ · V₂ ÷ c₁.<span>
V</span>₁(AgNO₃) = 0.250 M · 0.879 L ÷ 0.675 M.<span>
V</span>₁(AgNO₃) = 0.325 L · 1000 mL = 325 mL.

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