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
Answer:
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
Explanation:
not to certain if this is right or not.. but hope it helps!
Answer:
please further explain your question