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
An acidic solution is 0.1M in HCl and 0.2 H2so4. volume is equal to no of moles divided by molarity.
number of moles of HCl is 450ml x 0.1 divided by 1000 which is equal to 0.045 moles
volume of HCl is therefore 0.45 divided by 0.16 which is 2.81 litres
Number of moles of H2so4 is 450ml x 0.2 divided by 1000 which is equal to 0.09 moles
volume of H2SO4 IS 0.09 divided by 0.16 which is equal to 0.56 litres
Answer:
41.3 minutes
Explanation:
Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

So, fraction of original pressure = 
n here is number of half life
therefore, 
⇒ n= 3
it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.
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