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.

Answer 1

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