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17 April, 16:27

What volume (in L) of a 4.13M lithium nitrate solution would be needed to make 195mL of a 1.05M solution by dilution?

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  1. 17 April, 17:00
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    0.0496 L

    Explanation:

    You are diluting the LiNO₃ solution, so you can use the dilution formula.

    c₁V₁ = c₂V₂ Divide each side by c₁

    V₁ = V₂ * c₂/c₁

    c₁ = 4.13 mol·L⁻¹; V₁ = ?

    c₂ = 1.05 mol·L⁻¹; V₂ = 195 mL Calculate V₁

    V₁ = 195 * 1.05/4.13

    V₁ = 195 * 0.2542

    V₁ = 49.6 mL Convert to litres

    V₁ = 49.6 * 1/1000

    V₁ = 0.0496 L
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