Question

1. A student prepares two aqueous solutions. For each solution:

-Calculate the molarity.

-Write a chemical equation representing the formation of the solution. Be sure to include phase labels and to show any ions that dissociate.

-Draw a molecular picture of the solution showing any ions, solute molecules and water molecules.

A. 2.91 g Na₂CO₃ to make 25.0 mL solution.

B. 6.14 mL ethanol (d=0.789 g/mL) to make 75.0 mL of solution.

C. Which solution is more concentrated?

D. Identify each solution as a strong, weak, or non-electrolyte.

Answer 1

in case 1: 2.91g of Na₂CO₃   is taken to form 25ml (0.025 litre ) of soution .

moles of solute = given mass of solute / molecular mass (MM= 105.9g/moles)

2.91g /105.9 g moles^-1 =0.027 moles

Molarity = moles of solute / volume of solution in litres = 0.027mole / 0.025 litres = 1.098 mole/ litres or 1.098 M

In this reaction Na₂CO₃ in water to gives NaOH and CO₂ as products:

Na₂CO₃  + H₂O → 2Na⁺  (aq) +CO₃ ^2- and H₂O → H⁺ (aq)  + OH^- (aq)

2Na⁺ (aq) + 2 OH^- (aq) → 2NaOH (aq) and 2H⁺ (aq) + CO₃^2- → H₂O + CO₂  

Similarly in case of ethanol (molecular mass= 46 g/ moles) , mass of ethanol=6.14ml x 0.789 g/ ml = 4.48g (mass = volume x density)

Molarity = moles of solute / volume of solution in litres  (moles of solute = 4.84 g/ 46 g moles^-1 = 0.1052 moles)

= 0.1052 moles/ 0.075 litres= 1.404 moles/litre or 1.404 M

In this case ethanol and water show intermolecular hydrogen bonding as shown in picture

The solution of etanol is more concentrated ( 1.404 M) than solution of sodium carbonate ( 1.098)

Here Na₂CO₃ is strong electrolyte as it produce positive and negative ions in water whereas ethanol is weak electrolyte because it do not break into ions and form hydrogen bonding