Question

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:

How many mL of a 14.0 M stock solution must be used to make 250 mL of a 1.75 M solution?







You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?







100 mL of 6.00 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.50 M?







Water is added to 4.0 L of 6.0 M antifreeze until it is1.5 M. What is the total volume in L and mL) of the new solution?







If we have 1 L of 3.0 M HCl, what is the molarity of the solution if the acid is diluted to a volume of 6.0 L?

Answer 1

Molarity = moles of solute/Vol of solution in L

Ans.1

Mass of ethanol = 30g

Molar mass = 48.07g/mol

Moles = mass/molar mass = 30g/48.07g/mol = 0.624

Vol =6L

Molarity = 0.624mols/6L = 0.104M

Ans.2:

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

Ans.3:

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

Ans.4:

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