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8 February, 03:13

You need to make an aqueous solution of 0.196 M lead nitrate for an experiment in lab, using a 125 mL volumetric flask. How much solid lead nitrate should you add

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Answers (2)
  1. 8 February, 03:23
    0
    We have to ad 8.11 g of Pb (NO₃) ₂

    Explanation:

    To determine the mass of solute that would be needed for this experiment we have to think in the relation of Molarity and volume

    Molarity = Mol of solute / Volume (L)

    Therefore, Molarity. Volume (L) = Mol of solute

    We convert the volume of the volumetric flask in mL to L

    125 mL. 1L / 1000mL = 0.125 L

    0.196 mol/L. 0.125L = Moles of solute → 0.0245 moles

    Now we convert these moles, to mass

    Solute: Pb (NO₃) ₂ → Molar mass: 331.2 g/mol

    Moles. molar mass = mass → 0.0245 mol. 331.2 g/mol = 8.11 g
  2. 8 February, 03:41
    0
    We have to add 8.11 grams of lead nitrate

    Explanation:

    Step 1: data given

    Molarity of lead nitrate (Pb (NO3) 2) = 0.196 M

    Volume = 125 mL = 0.125 L

    Molar mass Pb (NO3) 2 = 331.2 g/mol

    Step 2: Calculate moles Pb (NO3) 2

    Moles Pb (NO3) 2 = volume Pb (NO3) 2 * molarity Pb (NO3) 2

    Moles Pb (NO3) 2 = 0.125 L * 0.196 M

    Moles Pb (NO3) 2 = 0.0245 moles

    Step 3: Calculate mass Pb (NO3) 2

    Mass Pb (NO3) 2 = moles Pb (NO3) 2 * molar mass Pb (NO3) 2

    Mass Pb (NO3) 2 = 0.0245 moles ¨331.2 g/mol

    Mass Pb (NO3) 2 = 8.11 grams

    We have to add 8.11 grams of lead nitrate
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