An aqueous solution is 36.9% by mass sodium hydroxide and has a density of 1.03 g mL What is the mole percent of NaOH?
The Correct Answer and Explanation is:
To calculate the mole percent of sodium hydroxide (NaOH) in a 36.9% by mass aqueous NaOH solution with a density of 1.03 g/mL, follow these steps:
1. Determine the mass of NaOH in the solution
The solution is 36.9% by mass NaOH, meaning 36.9 grams of NaOH are present in 100 grams of the solution. Therefore, in 100 grams of solution, we have:
- Mass of NaOH = 36.9 g
2. Determine the mass of water in the solution
The remaining mass of the solution is water. Since the total mass is 100 grams, the mass of water is:
- Mass of water = 100 g – 36.9 g = 63.1 g
3. Calculate the number of moles of NaOH
To find the number of moles of NaOH, use its molar mass. The molar mass of NaOH is:
- NaOH: 22.99 (Na) + 15.999 (O) + 1.008 (H) = 39.997 g/mol
Now, calculate the moles of NaOH:Moles of NaOH=Mass of NaOHMolar Mass of NaOH=36.9 g39.997 g/mol=0.922 mol\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar Mass of NaOH}} = \frac{36.9 \, \text{g}}{39.997 \, \text{g/mol}} = 0.922 \, \text{mol}Moles of NaOH=Molar Mass of NaOHMass of NaOH=39.997g/mol36.9g=0.922mol
4. Calculate the moles of water (H₂O)
The molar mass of water (H₂O) is:
- H₂O: 2 × 1.008 (H) + 15.999 (O) = 18.015 g/mol
Now, calculate the moles of water:Moles of water=Mass of waterMolar Mass of water=63.1 g18.015 g/mol=3.5 mol\text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar Mass of water}} = \frac{63.1 \, \text{g}}{18.015 \, \text{g/mol}} = 3.5 \, \text{mol}Moles of water=Molar Mass of waterMass of water=18.015g/mol63.1g=3.5mol
5. Calculate the total moles of the solution
The total number of moles in the solution is the sum of the moles of NaOH and the moles of water:Total moles=0.922 mol NaOH+3.5 mol water=4.422 mol\text{Total moles} = 0.922 \, \text{mol NaOH} + 3.5 \, \text{mol water} = 4.422 \, \text{mol}Total moles=0.922mol NaOH+3.5mol water=4.422mol
6. Calculate the mole percent of NaOH
The mole percent of NaOH is given by the ratio of the moles of NaOH to the total moles of the solution, multiplied by 100:Mole percent of NaOH=(Moles of NaOHTotal moles)×100=(0.9224.422)×100=20.85%\text{Mole percent of NaOH} = \left( \frac{\text{Moles of NaOH}}{\text{Total moles}} \right) \times 100 = \left( \frac{0.922}{4.422} \right) \times 100 = 20.85\%Mole percent of NaOH=(Total molesMoles of NaOH)×100=(4.4220.922)×100=20.85%
Final Answer:
The mole percent of NaOH in the solution is 20.85%.
