CHEM 1201 Lecture Notes - Lecture 11: Molar Mass, Sodium ChloridePremium
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Chem 1201 - Lecture 11 - Solution Calculations
● tells us how much of a compound or solute, in mol’s, is dissolved in the solvent, in liters,
and has the units of mol/L read as moles per liter.
● Molarity = moles (mol)
○ Calculate the molarity (M) of a solution prepared by dissolving 25.5 g CaCl2 (s)
(MM = 110.9 g/mole) in 250.0 mL. of water.
1. Find mol’s of CaCl2
■ Moles = (grams) (1)
■ Moles of CaCl2 = (25.5g) (mol)
■ Mol’s of CaCl2 = 0.23 mol
2. Find volume in liters (L). Convert 250 ml to L’s. Remember: 1000 ml = 1 L
■ Liters of solution = (250ml) (L)
■ Liters of solution = 0.250
3. Find Molarity
■ Molarity of CaCl2= (0.23 mol) = 0.92 M
What about the concentration of each ion? Stoich Conversion
● The symbols [ ] represent concentration in mol/L.
● Therefore given the [CaCl2] = 0.92 M. What is the [Ca+2] and [Cl- ]?
● (0.92 M of CaCl2) * (1 Ca+2) = 0.92 M of Ca+2
● (0.92 M of CaCl2) * (2 Cl-) = 1.84 M of Cl-
● GIVEN the Molarity (mol/L) of a particular solution, we can easily calculate:
1. the number of moles of a compound in solution given the volume of solution.
2. (and, by using molar mass) the mass of solute can then be determined by
multiplying mol’s times molar mass.
3. if seeking volume of solution; dividing molarity by mols gives you the volume in
liters. WATCHOUT!! You may need to convert grams to mol’s and liters to ml’s.
4. Summary Remember the rearrangement of units for molarity:
■ M = mol/L → M*L = mol → L = mol/M
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