Bonjour, pouvez vous m'aider s'il vous plait merci.... Pergunta de ideia dePikachu972

Pikachu972 @Pikachu972

May 2019 1 101 Report

Bonjour, pouvez vous m'aider s'il vous plait
merci

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scoladan

Verified answer

Bonjour,

1) M(AlCl₃) = M(Al) + 3xM(Cl) = 27 + 3x35,5 = 133,5 g.mol⁻¹

⇒ n(AlCl₃) = m(AlCl₃)/M(AlCl₃) = 668.10⁻³/133,5 ≈ 5,00.10⁻³ mol

2) C = n(AlCl₃)/Vsolution = 5,00.10⁻³/200,0.10⁻³ = 2,50.10⁻² mol.L⁻¹

3) D'après l'équation de dissolution :

[Al³⁺] = C = 2,50.10⁻² mol.L⁻¹

et [Cl⁻] = 3 x C = 7,50.10⁻² mol.L⁻¹

2 votes Thanks 0

Pikachu972 Merci mais d’ou sort la masse atomique M

scoladan M est la masse molaire. Voir livre de cours ou autre

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