Question

0.212 g of magnesium were dissolved in hydrochloric acid and the volume of hydrogen collected
over water at 16degreeC and 750 mm pressure, was 213.5 ml. Calculate the eq. wt. of the metal.

Answer 1

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Answer 2

Given info : 0.212 g of magnesium were dissolved in hydrochloric acid and the volume of hydrogen collected over water at 16°C and 750 mm pressure, was 213.5 ml.

To find : equivalent weight of the metal is ...

solution : first find no of moles of HCl in the given reaction.

volume of hydrogen, V = 213.5 ml = 0.2135 L

temperature, T = 16°C = 273 + 16 = 289 K

pressure = 750 mm Hg = 750/760 atm = 0.987 atm

now using formula, PV = nRT

⇒n = PV/RT

= (0.987 atm × 0.2135 L)/(0.082 atm.L/mol/K × 289K)

= 0.00889207952 ≈ 0.009 mol

magnesium is combined with hydrochloric acid as

Mg + 2HCl ⇒MgCl₂ + H₂

here you see 2 moles of HCl is used to form 1 mole of hydrogen gas.

so, no of moles of HCl to form 0.009 mol hydrogen gas = 2 × 0.009 = 0.018 mol

from titration,

moles of HCl = equivalents of Metal

⇒0.018 = weight of metal/(equivalent weight of metal)

⇒equivalent weight of metal = 0.212g/0.018

= 11.778 g

Therefore the equivalent weight of metal is 11.778 g.