What is the concentration of an alcl3 solution if 150. ml of the solution contains 550. mg of cl- ion?

The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M

calculation

concentration = moles /volume in liters

volume in liters = 150 /1000= 0.15 L

number of moles calculation

write the equation for dissociation of Al2Cl3

that is AlCl3 ⇔ Al^3+ + 3 Cl ^-

find the moles of Cl^- formed

moles =mass/molar mass

mass in grams= 550/ 1000 =0.55 grams

molar mass of Cl^- =35.5 g/mol

moles is therefore= 0.55/35.5 =0.0155 moles

by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles

concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M

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