A solution is made by dissolving 28.8 grams of aluminum chloride, AlCl_3, in eno

Question

A solution is made by dissolving 28.8 grams of aluminum chloride, AlCl_3, in enough water to make exactly 100 mL of solution. what is the concentration (molarity) of the following in mol/L? [AlCl_3=2.161M] (Al^3+=__M), [Cl^-=__M] **(i was able to calculate the molarity for the whole solution but how do you do the individual Al^3+ and Cl^-**

Explanation / Answer

AlCl3 dissociates in water as follows AlCl3 -----> Al+3 + 3Cl- so 1 mole of AlCl3 will give 1 mole of Al+3 ion and 3 moles of Cl- ions so firstly you calculate the number of moles of AlCl3 this is easy = mass/molecular mass = 28.8/(27 +3 x 35.5) = 2.161 M now since the volume of the solution remaons constant hence we can say as follows hence mole of Al+3 ion = 2.161 x 0.1 = 0.2161 moles and moles of Cl- ions = 2.161 x 3 x 0.1 = 0.6483 moles so molarity of Al+3 = moles/volume = 0.2161/0.1 = 2.161 M and molarity of Cl- = moles/volume = 0.6483/0.1 = 6.483 M hope this helps !

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.