Treatment of 1.385 g of an unknown metal, M, with an excess of aqueous HCl evolv

Question

Treatment of 1.385 g of an unknown metal, M, with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0°C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and the remaining HCl produces a white crystalline compound, MClx. After dissolving the compound in 25.0 g of water, the freezing point of the resulting solution was –3.53°C.

M (s) + HCl (aq) MClx (aq) + H2 (g)

D. What is the molality, m, of particles (ions) in the solution of MClx? Do not include units.

Explanation / Answer

Answer – We are given, freezing point of the resulting solution = –3.53°C

Mass of water = 25.0 g, Tf = 3.53°C

volume of H2 = 382.6 mL at 20.0°C and

P = 755 mm Hg /760 mm Hg = 0.993 atm

So, moles of H2 = PV/RT

                    = 0.993 atm * 0.3826 L / 0.0821 * 293 K

                  = 0.0158 moles

mass of H2 = 0.0158 moles * 2.016 g/mol

                  = 0.03185 g

Total mass of H2 come from the HCl so, mass of HCl

Moles of HCl = 2*0.0158 moles = 0.0316 moles * 36.46 g/mol

                       = 1.152 g

Law of Conservation of Mass:
(1.385 g M) + (1.152 g HCl) - (0.03185 g H2) = 2.505 g MClx produced

So, x = 2.505 g / 1.152 g = 2

So the MCl2 so total number ions = 3

So,

Tf = i*Kf*m

So, m = Tf / i*Kf

          = 3.53°C / 3*1.86°C.m-1

          =0.633 m

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