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What is the mass (in g) of K4Fe(CN)6 · 3H2O needed to prepare 100.00 mL of 0.0254 M solution?
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What is the mass (in g) of K4Fe(CN)6 · 3H2O needed to prepare 100.00 mL of 0.0254 M solution?

User SunLiWei
by
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1 Answer

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Molar mass

  • 368+3(18)
  • 368+54
  • 422g/mol

Now

  • molarity=moles of solute/vol of solution
  • 0.0254=n/0.1
  • n=0.00254mol

Now

  • mass=moles×molar mass
  • mass=0.00254×368
  • mass=0.9372g
User PeterA
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8.4k points
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