5. The argentometric gravimetric analysis of an alkaline earth metal chloride wa

Question

5. The argentometric gravimetric analysis of an alkaline earth metal chloride was conducted as usual (by dissolving the pure metal chloride in water, precipitating silver chloride, coagulating, filtering, and drying the colloid, and finally taking the mass of the solid.) Three samples of the metal chloride weighing 0.372, 0.365, 0.383 grams were analyzed giving 0.512, 0.503, 0.527 grams of silver chloride respectively. (That is, the first number in each set is one sample, the second is the next and the third number in each set is the third sample.)

Calculate the % chloride in the metal chloride samples. {Gravimetric Factor = 0.2474} B.)

Which alkaline earth metal chloride is this? (Show how you determined your answer.)

Explanation / Answer

From the given data -

The number of mols of AgCl =0.512 g/143.5 g/mol = 0.00356 mol (therefore same number of mols of Cl should be present in the sample)

Thus thenumber of mols of Metal chloride in original sample = 0.00356/2 = 0.00178 mol

From the gravimetric factor, 1 g of metal chloride contains 0.2474 g of Cl (0.00696 mol)

Therefore, the mass of metal chloride to have 0.0178 mols of Cl = 0.00356/0.00696 = 0.512 g

Thus molar mass of Metal chloride = 0.512 g / 0.00178 mol = 287.64 g/mol

Since , the formula of Metal chloride is MCl2 ,

The molar mass of metal = 287.64 - 2(35.5) = 216.64 g/mol (Metal)

This indicates that the metal must be Radon, Rn

The % chloride = (0.2474 / 1 )*100 = 24.74 %

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