The ksp of salt agcl at 25 degree celsius is 2.56 10^-10.Then how much volume of water is required to dissolve .01 mole of salt.
Answers
625 L of water is required to dissolve .01 mole of AgCl salt.
Explanation:
=> It is given that,
The ksp of AgCl = 2.56 * 10⁻¹⁰
Mole of AgCl salt = 0.01 mole
=> Decomposition reaction of AgCl in water:
AgCl => Ag⁺(ₐq) + Cl⁻(ₐq)
0.01 mole of AgCl gives 0.01 mole of Ag⁺ and 0.01 mole of Cl⁻
=> Thus, Solubility of both Ag⁺ and Cl⁻ is equals to s, then ksp:
Ksp = [Ag⁺]*[Cl⁻] = s * s = s² = 2.56 * 10⁻¹⁰
s = √2.56 * 10⁻¹⁰
= 1.6 * 10⁻⁵ mol/L
=> Volume of water required to dissolve .01 mole of salt:
s = mole / Volume (L)
Volume (L) = mole / s [mole of Ag⁺ and Cl⁻ is equals to the mole of AgCl]
= 0.01 / 1.6 * 10⁻⁵
= 0.00625 * 10⁵
= 625 L
Thus, 625 L of water is required to dissolve .01 mole of salt.
Learn more:
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Explanation:
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