moles and solutions

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Created by
mia

Cards (10)

  • what is molarity?

    measure of concentration
  • what is the concentration calculation?
    concentration = moles / volume
  • what is a dilution calculation?
    adding water to a solution causes the concentration to decrease
    water doesn't react with the solution
    moles don't change
  • example: 100cm^3 of 0.125moldm^-3 MgCl2 is added to 50cm^3 of 0.65moldm^-3 NH4Cl. calculate the overall concentration of Cl- ions in the mixture
    moles of MgCl2 - 0.1 x 0.125 = 0.0125 moles
    moles of Cl in MgCl2 - 0.0125 x 2 = 0.025 moles
    moles of NH4Cl - 0.05 x 0.65 = 0.0325 moles
    total moles of Cl - 0.025 + 0.0325 = 0.0575 moles
    total volume of Cl - 0.1 + 0.05 = 0.15dm^3
    concentration of Cl- ions - 0.0575 / 0.15 = 0.383moldm^-3
  • what is a standard solution?

    solution that we know the concentration of
  • what is a titration?

    method by which a solution of an acid is accurately added to a known volume of a base until neutralisation just occurs
  • what is a titration also known as?

    volumetric analysis
  • example: 12.3g of HX is dissolved in 250cm^3 of water. a 25cm^3 sample of solution requires 24cm^3 of 0.4 moldm^-3 NaOH for neutralisation. calculate the Mr of the acid and identify X
    HX +NaOH --> NaX + H2O
    moles of NaOH - 0.024 x 0.4 = 0.0096 moles
    moles of HX in 25cm^3 - 0.0096 x 1 = 0.0096 moles
    moles of HX in 250cm^3 - 0.0096 x 10 = 0.096 moles
    Mr of HX - 12.3 / 0.096 = 128.1
    Mr of X - 128.1 - 1 =1 27.1
    X is Iodine
  • what is a back titration?
    concentration of unknown is determined by reacting it with known amount of excesss reagent
    remaining is titrated with second reagent to show how much of excess reagent was used in initial reaction
  • example: 4.06 g of impure MgO was completely dissolved in 100cm³ of HCl, of concentration 2moldm-3 (in excess). Excess HCl required 19.7cm³ of NaOH (0.2moldm-3) for neutralisation. Calculate percentage purity of MgO
    1. MgO + 2HCl --> MgCl2 + H2O
    2. HCl + NaOH --> NaCl + H2O
    moles of HCl (1.) - 0.1 x 2 =0.2 moles
    moles of excess HCl reacted - 0.0197 x 0.2 = 0.00394 moles
    moles of HCl reacting with MgO - 0.2 - 0.00394 = 0.19606 moles
    moles of MgO - 0.19606 / 2 =0.09803 moles
    mass of MgO - 0.09803 x 40 = 3.95g
    percentage purity of MgO - 4.06 / 3.95 = 97.3%