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					           IB Topic 1: Quantitative Chemistry
           1.5 Solutions

   Distinguish between the terms solute, solvent, solution
    and concentration (g dm-3 and mol dm-3)
   Solve problems involving concentration, amount of
    solute and volume of solution.
              Distinguish between the terms solute, solvent,
              solution and
              concentration (g dm-3 and mol dm-3)
Solution: Homogeneous mixtures of two or
    more substances. Most common is
    solid, liquid or gas dissolved in a liquid
    (usually water). These are called
    aqueous solutions. Can have other
    solutions such as solid-solid (alloy) or
    gas-gas (air).
Solute: The dissolved particles. Usually the
    substance in the least amount.
Solvent: The dissolving medium. Usually
    the substance in the greater amount.
               Distinguish between the terms solute, solvent,
               solution and
               concentration (g dm-3 and mol dm-3)
         Properties of Solutions
               Solubility

Solubility is the amount of solute that dissolves
     in a given amount of solvent at a given
     temperature to produce a saturated
     solution. Units: grams solute/100 g solvent
NaCl: solubility of 36.2 g/ 100 g water at 25 oC
Any amount less than that is an unsaturated
     solution.
A solution that contains more solute than it
     should theoretically is supersaturated.
             Distinguish between the terms solute, solvent,
             solution and
             concentration (g dm-3 and mol dm-3)
Concentration: Measure of the amount of solute dissolved
   in a given amount of solvent.

   g dm-3 (g/dm3)
       Is the number of grams of solute dissolved per dm3 of solution


   mol dm-3 (mol/dm3) Molarity (M)
       is the number of moles of solute dissolved per dm3 of solution.
       M = mol dm-3.
       Use [ ] to signify concentration in molarity.
              Solve problems involving concentration,
              amount of solute and volume of solution.

Find the concentration in g dm-3 and mol dm-3 of a solution containing
     2.00 g sodium hydroxide in 125 cm3 of solution.

g dm-3
   125 cm3 = 0.125 dm3
   2.00 g/0.125 dm3 = 16.0 g dm-3


mol dm-3
   2.00 g NaOH = 2.00 g/ 40.01 g mol-1 = .0500 mol
   .0500 mol/0.125 dm3 = 0.400 mol dm-3
           Solve problems involving concentration,
           amount of solute and volume of solution.

Calculate the amount of hydrochloric acid (in mol & g)
    present in 23.65 cm3 of 0.100 mol dm-3 HCl(aq)?
           Solve problems involving concentration,
           amount of solute and volume of solution.

Calculate the amount of hydrochloric acid (in mol & g)
    present in 23.65 cm3 of 0.100 mol dm-3 HCl(aq)?

Molarity = mol dm-3 so Molarity x dm3 = mol

mol = 0.100 mol dm-3 x .02365 dm-3 = .002365 mol HCl

grams = .002365 mol x 36.46 g mol-1 = .0862 g HCl
          Solve problems involving concentration,
          amount of solute and volume of solution.

What volume of a 1.25 mol dm-3 potassium permanganate
   solution, KMnO4(aq), contains 28.6 grams KMnO4?
           Solve problems involving concentration,
           amount of solute and volume of solution.

What volume of a 1.25 mol dm-3 potassium permanganate
   solution, KMnO4(aq), contains 28.6 grams KMnO4?

Molarity = mol dm-3 so dm-3 = mol/Molarity

mol = 28.6 g / 158.04 g mol-1 = .181 mol

dm-3 = .181 mol/1.25 mol dm-3 = .145 dm3
           Solve problems involving concentration,
           amount of solute and volume of solution.

What will be the concentration of the solution formed by
   mixing 200 cm3 of 3.00 mol dm-3 HCl(aq) with 300 cm3
   of 1.50 mol dm-3 HCl(aq)?
         Solve problems involving concentration,
         amount of solute and volume of solution.

What will be the concentration of the solution formed by
    mixing 200 cm3 of 3.00 mol dm-3 HCl(aq) with 300 cm3
    of 1.50 mol dm-3 HCl(aq)?
Find total moles
    .200 dm-3 x 3.00 mol dm-3 = .600 mol HCl
    .300 dm-3 x 1.50 mol dm-3 = .450 mol HCl
    Total moles = 1.050 mol
Find total volume: .200 dm3 + .300 dm3 = .500 dm3
Concentration: 1.050 mol/ .500 dm3 = 2.10 mol dm-3
Terms to Know
   Element                  Molar ratio
   Compound                 Ionic equation
   Empirical formula        Solid
   Molecular formula        Liquid
   Structural formula       Gas
   Percent composition      Ideal gas equation
   Percentage yield         Molar volume of a gas
   Molar mass               Solute
   Mole                     Solvent
   Avogadro’s constant      Solution
   Chemical equation        Concentration

				
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posted:9/27/2012
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