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if 45.55ml of 0.0105M EDTA solution was needed to titrate the Mg in a 0.2600g sample...

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Chemistry

if 45.55ml of 0.0105M EDTA solution was needed totitrate the Mg in a 0.2600g sample of the mineral hydromagnesite,3MgCO3Mg (OH)2• 3H2O (365.3 g/mol), what was the weight percent ofhydromagnesite in the sample? what was the weight percent ofMg+2 inthe sample.

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4.4 Ratings (649 Votes)

Solution :-

lets first calculate the moles of the EDTA

moles of EDTA = 0.0105 mol per L * 0.04555 L = 0.0004783 mol

mole ratio of the EDTA to Mg2+ is 1 : 1

so the moles of Mg^2+ = 0.0004783 mol

mass of Mg^2+ = moles * molar mass

                        =0.0004783 mol * 24.305 g per mol

                       = 0.01163 g Mg

% of Mg^2+ = (mass of Mg^2+ / mass of sample )*100 %

                     = (0.01163 g / 0.2600 g)*100%

                     = 4.47 %

now l;ets calculate the weight percent of the Mg in the sample

% Mg = ((4*24.305 g)/365.3 g)*100%

         = 26.61 %
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