Calculate the molarity of a solution containing 30.0 g of ethanol (C2H5OH) in 6.

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?










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Explanation / Answer

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

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