Calculate the molarity of a solution containing 30.0 g of
ethanol (C2H5OH) in 6.00 L of solution.
What is the molarity of a solution containing 2.00 g of NaOH in 1000 mL of solution?
How many grams of sodium carbonate (Na2CO3) are needed to prepare 2.50 L of a 0.15 M solution?
How many grams of KOH are needed to prepare 100 mL of a 0.50 M solution?
Lab 9 Worksheet: Concentration of Solutions (Ch4) Name:
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Molarity = moles of solute/Vol of solution in L
Mass of ethanol = 30g
Molar mass = 48.07g/mol
Moles = mass/molar mass = 30g/48.07g/mol = 0.624
Molarity = 0.624mols/6L = 0.104M
Mass of NaOH = 2g
Molar mass = 39.99g/mol
Moles = 2g/39.99g/mol = 0.05mol
Vol = 1000ml = 1L
Molarity = 0.05mol/1L = 0.05M
Volume = 2.50L
Molarity = 0.15M
Molarity= moles/Volume in L
Moles = Molarity * Volume
Moles = 0.15*2.50 = 0.375mols
Molar mass of Na2CO3 = 105.98g/mol
Moles = mass/molar mass
So,mass of Na2CO3 = moles* molar mass =0.375moles*105.98g/mol =39.74g
MAss of Na2CO3 = 39.74g
Molarity = 0.5M
Volume = 100ml = 0.1L
Moles = molarity*vol = 0.5*0.1 = 0.05moles
Molar mass of KOH = 56.11g/mol
Mass of KOH = moles* molar mass =0.05mol*56.11g/mol = 2.81g
Mass of KOH= 2.81g
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