Question

12. What is the molarity of 0.31 mol of NaCl in enough water to make 2.0 L of solution?

13. What is the molarity of the nitrate ions when 0.60 mol of barium nitrate, Ba(NO₃)₂, is dissolved in enough water to make 2.0 L of solution?

14. Calculate the molarity of calcium hydroxide that has been prepared by dissolving 2.24 g of calcium hydroxide in water and diluting the solution to a final volume of 4.00 L.

15. Calculate the molarity of the hydroxide ions in a solution that has been prepared by dissolving 70.3 g of barium hydroxide (Ba(OH)₂) in water and diluting the solution to a final volume of 4.00 L. (M.W's: Ba 137.3 g/mol., O 16.0 g/mol, and H 1.0 g/mol)

16. How many moles of H⁺ are in 181 mL of 0.71 M nitric acid, HNO₃?

Answer 1

12-

Molarity is the measure of concentration of a solution, which is the number of moles of solute present per 1L volume of solution

i.e Molarilty = mols of solute / volume eof solution in Liter

Now given

here our mols of solute (NaCl) taken = 0.31 mol

Total volume of solution formed =  2.0 L

So Molarilty = mols of solute / volume eof solution in Liter

= 0.31 mol / 2.0 L

= 0.155‬ mol/L

= 0.155‬ M

13-

A Ba(NO₃)₂ dissociate in water as-

Ba(NO₃)₂ -------------> Ba⁺² + 2NO₃^-  

i.e we get two mols of nitrate ions (NO₃^-) from the dissolution of 1 mole of Ba(NO₃)₂

Now given we have taken Ba(NO₃)₂ = 0.60 mol

So mols of  nitrate ions produced from it = 0.60 mol * 2

= 1.2 mols

That means in our 2.0 L solution, we have now 1.2 mols of  nitrate ions

So

Molarilty of nitrate ions = mols of nitrate ions / volume of solution in Liter

= 1.2 mol / 2.0 L

= 0.60 mol/L

   = 0.60‬ M

14-

Given mass of calcium hydroxide (Ca(OH)2) taken =  2.24 g

So mols of Ca(OH)₂ taken = mass / molar mass of Ca(OH)₂

= 2.24 g / 74 g/mol

= 0.0302 mols

Molarilty of Ca(OH)₂ = mols of Ca(OH)₂ / volume of solution in Liter

= 0.0302 mol / 4.0 L

= 0.00756 mol/L

   = 7.56 * 10^-3‬ M

15-

1 mole of Ba(OH)₂ dissociate in water as-

Ba(OH)₂ -------------> Ba⁺² + 2OH^-  

i.e we get two mols of (OH^-) from the dissolution of 1 mole of Ba(OH)₂

Now given we have taken mass of Ba(OH)₂ = 70.3 g

Again mass of 1 mole of Ba(OH)₂ = (1*mass of Ba) + (2*mass of O) + (2*mass of H)

= (1*137.3 g/mol) + (2*16.0 g/mol) + (2*1.0 g/mol)

= (137.3 g/mol) + (32.0 g/mol) + (2.0 g/mol)

= 171.3 g/mol

So

mols of Ba(OH)₂ taken = mass / molar mass of Ba(OH)₂

= 70.3 g / 171.3 g/mol

= 0.41 mols

So mols of  OH^- produced from it = 0.41 mol * 2

= 0.82 mols

That means in our 4.0 L solution, we have now 0.82 mols of  hydroxide ions

So

Molarilty of hydroxide ions = mols of hydroxide ions / volume of solution in Liter

= 0.82 mol / 4.0 L

= 0.205 mol/L

   = 0.205‬ M