Molarity Problems Molarity = mol/L Molarity = moles of solute / Liters of solution Concentration ~How much solute is present in a solution compared to the solvent. Molarity (M)- moles of solute per liter of solution. M = mol/L 2.1 M AgNO means 2.1 mol of AgNO for 3 3 every one liter of solution Other measures of concentration Name
Abbrev. What it is molality m mol solute/kg solvent parts per million ppm g solute/g solvent x 106
parts per billion ppb g solute/g solvent x 109 mole fraction x mol solute/mol solution percent by mass % g solute/g solution x 100
percent by volume % mL solute/mL solvent x100 Molarity problems How many moles of HCl are in 125 mL of 2.5 M HCl? 2.5 mol HCl 1 L of soln. .125 L of soln = .31 mol HCl
Here we go What concentration solution would be prepared if 39 g of Ba(OH)2 were mixed in a 450 mL solution? 39 g Ba(OH)2 1 mol Ba(OH)2 =.2276 mol Ba(OH)2 171.316 g Ba(OH)2 M = mol/L .2276 mol Ba(OH)2 =.51 M Ba(OH)2
.45 L of solution More For a lab in this chapter, I need to make .60 L of 3.0 M NaOH, what mass of NaOH did I need? .6 L x 3.0 M NaOH = 1.8 mol NaOH 1.8 mol NaOH x 39.998 g/mol = 72 g NaOH More molarity problems How many moles of HCl are in 525 mL of 3.5 M HCl?
3.5 mol HCl 1 L of soln. .525 L of soln = 1.8 mol HCl Still going What concentration solution would be prepared if 48 g of Ca(OH)2 were mixed in a 450 mL solution? 48 g Ba(OH)2 1 mol Ba(OH)2
=.65 mol Ba(OH)2 74.096 g Ba(OH)2 M = mol/L .65 mol Ba(OH)2 .45 L of solution =1.4 M Ba(OH)2 For a lab in this chapter, I need to make .75 L of 3.0 M NaOH, what mass of NaOH did I need? .75 L x 3.0 M NaOH = 2.25 mol NaOH 2.25 mol NaOH x 39.998 g/mol = 90. g NaOH
Molarity Problems A 0.24 M solution of Na2SO4 contains 0.36 moles of Na2SO4. How many liters were required to make this solution? 0.36 mol Na2SO4 1L soln 0.24 mol = 1.5 L Na2SO4 Getting tougher 2 AgNO 3
+ BaCl2 2 AgCl + Ba(NO3)2 Balance the equation. If 1.2 L of .50 M AgNO3 is reacted completely, what molarity solution of Ba(NO3)2 will be created if the volume increased to 1.5 L? 1.2 L x .5 M AgNO3 = .6 mol AgNO3 . 6 mol AgNO3 1 mol Ba(NO3)2
2 mol Ag NO3 = .3 mol Ba(NO3)2 1.5 L = .20 M Ba(NO3)2 Last one 2 HNO 3 + Zn H2 + Zn(NO3)2 If you have .65 L of 1.2 M HNO3 and you react it completely what volume of H 2 gas will you produce at STP?
1.2 M HNO3 x .65 L = .78 mol HNO3 . 78 mol HNO3 .39 mol H2 1 mol H2 2 mol HNO3 22.4 L at STP 1 mol H2 =.39 mol H2 = 8.7 L at STP Homework Problem HNO3 + Zn H2 + Zn(NO3)2
If you have .45 L of 2.1 M HNO3 and you react it completely what volume of H 2 gas will you produce at STP? Solve for moles of H , then use Avogadros 2 law to get a volume at STP.
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