J.R. S. answered 02/06/21
Ph.D. University Professor with 10+ years Tutoring Experience
0.551 M AlCl3 = 0.551 moles AlCl3 / L
0.551 moles AlCl3 x 3.20 moles of particles / mole AlCl3 = 1.76 moles of particles
Concentration of particles = 1.76 M
Moh D.
asked 02/06/21AlCl₃ has a van't Hoff factor of i = 3.20. What is the concentration of particles in a 0.551 M solution of AlCl₃?
J.R. S. answered 02/06/21
Ph.D. University Professor with 10+ years Tutoring Experience
0.551 M AlCl3 = 0.551 moles AlCl3 / L
0.551 moles AlCl3 x 3.20 moles of particles / mole AlCl3 = 1.76 moles of particles
Concentration of particles = 1.76 M
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