Weird: number of moles of NaOH in 20 ml of 1M NaOH

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Discussion Overview

The discussion revolves around the calculation of the number of moles of NaOH in a 20 ml solution of 1M NaOH, exploring both the straightforward molarity approach and a density-based approach. Participants examine the implications of using density in this context and the definitions of molarity.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants calculate the number of moles in 20 ml of 1M NaOH as 20 mmol using the formula for molarity.
  • Others introduce a density approach, suggesting that 20 ml of a 1M NaOH solution, with a density of 1.04 g/ml, results in a mass of 20.8 g, leading to a calculation of 520 mmol when divided by the molar mass of NaOH.
  • One participant points out that the density of the solution primarily reflects the water content, not the NaOH solute itself.
  • Another participant emphasizes that the density approach is flawed because it uses the total mass of the solution rather than the mass of the solute.
  • It is noted that molarity is defined as moles of solute per volume of solution, which means the density is not necessary for this calculation.
  • Some participants argue that if one were to use the density approach, it would require determining the volume of NaOH specifically in the solution, rather than the volume of the entire solution.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the molarity calculation but disagree on the appropriateness of using the density approach, with some asserting it leads to incorrect conclusions.

Contextual Notes

There is a lack of consensus on how to properly apply the density approach, with unresolved questions about the volume of solute versus the volume of the entire solution.

p3t3r1
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Hi there, this is a not a homework question but rather question about the theory behind this problem.

Say if someone asks you to calculate the number of moles in 20 ml of a 1M NaOH solution, I would usually say it is going to be 20 mmol as you do 20 ml/1000 ml/L x 1M = 0.02 moles = 20 mmol.

However, if you consider the density approach, 1M of NaOH has a density of 1.04 g/ml so 20 ml is 20 x 1.04 g = 20.8 g/40 g/mol (MW of NaOH) that is going to yield 520 mmol of NaOH in 20 mll of 1 M NaOH.

Obviously the density approach give a number that makes no sense. I am just wondering why the density approach is not correct.

Cheers.
 
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1.04 g/ml is the density of the entire solution, most of which is water not NaOH.
 
p3t3r1 said:
Hi there, this is a not a homework question but rather question about the theory behind this problem.

Say if someone asks you to calculate the number of moles in 20 ml of a 1M NaOH solution, I would usually say it is going to be 20 mmol as you do 20 ml/1000 ml/L x 1M = 0.02 moles = 20 mmol.

However, if you consider the density approach, 1M of NaOH has a density of 1.04 g/ml so 20 ml is 20 x 1.04 g = 20.8 g/40 g/mol (MW of NaOH) that is going to yield 520 mmol of NaOH in 20 mll of 1 M NaOH.

Obviously the density approach give a number that makes no sense. I am just wondering why the density approach is not correct.

Cheers.

You are going wrong there, you shouldn't divide by the MW of NaOH. The 20.8g is the weight of the solution, so it makes no sense.
 
Say if someone asks you to calculate the number of moles in 20 ml of a 1M NaOH solution, I would usually say it is going to be 20 mmol as you do 20 ml/1000 ml/L x 1M = 0.02 moles = 20 mmol.

Uncomplicated! You are given 1 Molar NaOH. This is in moles per liter.

However, if you consider the density approach, 1M of NaOH has a density of 1.04 g/ml so 20 ml is 20 x 1.04 g = 20.8 g/40 g/mol (MW of NaOH) that is going to yield 520 mmol of NaOH in 20 mll of 1 M NaOH.

The density is not needed because you already are given the concentration is in moles per liter, and your question was for some number of milliliters; in this case, 0.020 liters. The given and the question do not involve density.
 
As the other posters said, you cannot use the density or volume of the entire solution. If you wanted to take the density approach, you would need to figure out the volume of NaOH in your solution, and multiply that by the density of solvated NaOH. That would give you the mass (which will be much smaller than your 20.8 g), which you could then divide by the molar mass.

The molarity calculations work because molarity is defined as moles SOLUTE/volume SOLUTION, so when you multiply by the volume of a solution it will cancel. In your calculations above, your units will not cancel because you are doing
20.8 g SOLUTION
-----------
40 g NaOH/mol NaOH
 

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