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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Answer:- C. 16.4 L
Solution:- The given balanced equation is:

From this equation, there is 2:1 mol ratio between HCl and hydrogen gas. First of all we calculate the moles of hydrogen gas from given grams of HCl using stoichiometry and then the volume of hydrogen gas could be calculated using ideal gas law equation, PV = nRT.
Molar mass of HCl = 1.008 + 35.45 = 36.458 gram per mol
The calculations are shown below:

= 
Now we will use ideal gas equation to calculate the volume.
n = 0.672 mol
T = 25 + 273 = 298 K
P = 101.3 kPa = 1 atm
R = 
PV = nRT
1(V) = (0.672)(0.0821)(298)
V = 16.4 L
From calculations, 16.4 L of hydrogen gas are formed and so the correct choice is C.