In a solution of NiCl2, the Cl concentration is 0.08684 M. How many grams of NiCl2 are in 338 mL of this solution?
2007-01-17 12:58:13 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Let the solution be called s. Atomic weights: Ni = 59 Cl = 35.5 NiCl2 = 130
338mLs x 0.08684molNi/1000mLs x 130gNiCl2/1molNiCl2 = (338)(0.08684)(130)/(1000) = 3.82gNiCl2
The 338mL is given. The first factor comes from the molarity. The mL solution cancel, leaving mol NiCl2. The second factor comes from the molecular (formula) weight. The mol NiCl2 cancel, leaving g NiCl2.
2007-01-17 13:12:09 · answer #1 · answered by steve_geo1 7 · 1⤊ 0⤋
2Cl + 1Ni --> 1NiCl2
well, there are half as many NiCl2's than there are Cl's, so divide Cl's molarity by two two get: 0.04342. it's in 0.338 L, so multiply to get the moles:
0.014676 moles. then multiply by the molar mass (129.6g) to get:
1.902 grams of NiCl2
2007-01-17 21:06:21 · answer #2 · answered by car of boat 4 · 0⤊ 0⤋