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25 February, 18:28

Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution is needed to form 0.525 g of silver chloride

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  1. 25 February, 19:19
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    1.22 mL

    Explanation:

    Let's consider the following balanced reaction.

    2 AgNO₃ + BaCl₂ ⇄ Ba (NO₃) ₂ + 2 AgCl

    The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:

    0.525 g * (1 mol/143.32 g) = 3.66 * 10⁻³ mol

    The molar ratio of AgCl to BaCl₂ is 2:1. The moles of BaCl₂ are 1/2 * 3.66 * 10⁻³ mol = 1.83 * 10⁻³ mol.

    The volume of 1.50 M barium chloride containing 1.83 * 10⁻³ moles is:

    1.83 * 10⁻³ mol * (1 L/1.50 mol) = 1.22 * 10⁻³ L = 1.22 mL
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