Assume you have 100. mL of the solution. Find the concentration of NaOH expresse

Question

Assume you have 100. mL of the solution. Find the concentration of NaOH expressed as: Hint: Start Here: 11. An aqueous solution of lye (NaOH) is 12.50 % ww Na Oll with a density of 11085 gm Then Convert: Mass of NaOH W31% Moles of NaOH Volume of Solution Mass of Solution Mass of Water Mole Fraction (NaOH) Moles of Water Mole Fraction (Water) Total Moles 12. A 15.0 mL of a 1.25 M solution of battery acid (H SOa) is diluted to 250. mL with fresh DI water. The diluted solution had an additional 150.0 mL of DI water added. Calculate the final concentration of the solution.

Explanation / Answer

A 12.5% w/v NaOH means that 12.5 grams of NaOH is dissolved per 100ml of solution.

1. Mass of NaOH

Since the amount of NaOH dissolved in 100ml of the solution is 12.5 grams and the volume of solution given is 100ml.

Mass of NaOH = 12.5 grams

2. Moles of NaOH

Molar mass of NaOH = Molar mass of Na + Molar mass of O + Molar mass of H

= (23+16+1)g/mol = 40 g/mol

Mass of NaOH = 12.5g

Now, Moles of NaOH = (Mass of NaOH)/(Molar mass of NaOH)

Moles of NaOH = 12.5/40 = 0.3125 mol

3. Volume of solution

It is given,

Volume of solution = 100ml

4. Mass of solution

We know that,

Density = Mass/Volume

So, Mass = Density*Volume

Now, Density of solution = 1.1085g/ml

Volume of solution = 100ml

So, Mass of solution = 1.1085*100

Mass of solution = 110.85g

5. Mass of water

  Mass of NaOH + Mass of water = Mass of solution

  Mass of water = Mass of solution - Mass of NaOH

  Mass of water = 110.85-12.5

  Mass of water = 98.35g

6. Moles of water

Molar mass of H2O = Molar mass of H2 + Molar mass of O

= (2+16)g/mol = 18 g/mol

Mass of H2O = 18g

Now, Moles of H2O = (Mass of H2O)/(Molar mass of H2O)

Moles of H2O? = 98.35/18 = 5.46mol

7. Total moles

Total moles = Moles of NaOH + Moles of H2O

Total moles = 0.3125+5.46

Total moles = 5.7725 mol

8. Converting to w/w%

w/w% = {(Weight of NaOH)/(Weight of solution)}*100

= (12.5/110.85)*100 = 11.27%

9. w/v%

It is given in the question,

w/v% = 12.50%

10. M (Molarity)

Molarity = (Moles of NaOH)/(Volume of solution in liters)

Moles of NaOH = 0.3125

Volume of solution in liters = 100/1000 = 0.1 L

Molarity = 0.3125/0.1 = 3.125M

11. m (Molality)

Molality = (Moles of NaOH)/(Weight of water in kg)

Moles of NaOH = 0.3125mol

Weight of water in kg = 98.35/1000 = 0.09835kg

Molality = 03125/0.09835 = 3.177mol/kg

12. Mole Fraction (NaOH)

Mole Fraction of NaOH = (Moles of NaOH)/(Total Moles)

= 0.3125/5.7725 = 0.0541

13. Mole Fraction (Water)

Mole Fraction of NaOH + Mole Fraction of Water = 1

Mole Fraction of Water = 1 -  Mole Fraction of NaOH

= 1 -  0.0541 = 0.9459

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