0.2590 g of AlCl₃ contains 0.0058 mole of chloride ion, Cl¯.
We'll begin by obtaining the mass of 1 mole of AlCl₃. This can be obtained as follow:
1 mole of AlCl₃ = 27 + (3×35.5)
= 27 + 106.5
= 133.5 g
Next, we shall write the balanced dissociation equation for AlCl₃. This is illustrated below:
AlCl₃(aq) —> Al³⁺(aq) + 3Cl¯
From the balanced equation above,
133.5 g of AlCl₃ contains 3 moles of Cl¯
Finally, we shall determine the number of mole of chloride ion, Cl¯ in 0.2590 g of AlCl₃. This can be obtained as follow:
From the balanced equation above,
133.5 g of AlCl₃ contains 3 moles of Cl¯.
Therefore,
0.2590 g of AlCl₃ will contain = (0.2590 × 3) / 133.5 = 0.0058 mole of Cl¯.
Thus, 0.2590 g of AlCl₃ contains 0.0058 mole of chloride ion, Cl¯.
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