Berikut ini adalah pertanyaan dari masturahnazifa pada mata pelajaran Kimia untuk jenjang Sekolah Menengah Atas
Calculate the mass of precipitate formed.
Jawaban dan Penjelasan
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Jawaban:
To solve this problem, we need to use the fact that a mole of a substance is a unit of mass equal to the atomic or molecular weight of that substance in grams.
We know that the concentration of the sodium sulphate solution is 0.5 mol/dm³, and that we have added 10 cm³ of the solution. Therefore, the number of moles of sodium sulphate in the solution is 0.5 x (10/1000) = 0.005 moles.
Sodium sulphate has a molecular weight of 142.04 g/mol, so the mass of sodium sulphate in the solution is 0.005 x 142.04 = 0.7102 g.
When sodium sulphate is added to lead (II) nitrate, a white precipitate of lead (II) sulphate is formed. The formula for lead (II) sulphate is PbSO4, and its molecular weight is 303.34 g/mol.
Thus, the mass of lead (II) sulphate precipitate formed is 0.005 x 303.34 = 1.5167 g.
Therefore, the mass of precipitate formed is 1.5167 grams.
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Last Update: Sat, 01 Apr 23