How Do You Calculate Grams of Solute for Specific Solutions?

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Discussion Overview

The discussion revolves around calculating the grams of solute needed to prepare specific solutions, focusing on a 0.1 N solution of Ca(OH)2 and a 5% NaCl solution. It includes aspects of normality, molarity, and the implications of different concentration definitions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant attempts to calculate the grams of Ca(OH)2 needed for a 0.1 N solution, suggesting a conversion based on the number of hydrogen ions.
  • Another participant proposes preparing a 0.2 M solution of Calcium Hydroxide, which raises questions about the correct molarity for the solution.
  • A participant clarifies that a 0.05 M solution of Ca(OH)2 corresponds to 0.05 M in Ca2+ and 0.1 M in OH-, referencing the definition of normality.
  • There is uncertainty regarding whether the 5% NaCl solution is w/v or w/w, with implications for the calculation of grams needed.
  • Different approaches to calculating the mass of NaCl are discussed, including a simple method assuming a density of 1 g/mL and a more precise method using density tables.
  • One participant seeks clarification on setting up the calculation for the NaCl solution, indicating a need for further guidance.

Areas of Agreement / Disagreement

Participants express differing views on the correct molarity for the Ca(OH)2 solution and the method for calculating the grams of NaCl needed, indicating that multiple competing views remain unresolved.

Contextual Notes

There is ambiguity regarding the definitions of concentration (w/v vs. w/w) for the NaCl solution, which affects the calculations. Additionally, the assumptions made about the density of solutions may influence the accuracy of the results.

azncocoluver
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Homework Statement


Calculate how many grams of solute are necessary to prepare:
1) 100 mL of 0.1 N (Normal Solution) Ca(OH)2 (molecular weight = 74)

2) 100 mL 5% NaCl (molecular weight = 40)


Homework Equations


Molarity = # of moles / # of liters


The Attempt at a Solution


1. I set up dimensional analysis.

74g Ca(OH)2 divided by 2 since there are 2 Hydrogens = 37g Ca(OH)2 for Molecular Weight

100ml * (1 L / 1,000 mL) * (.05 mol / 1 L) * (37g / 1 mol) = 0.185g

I'm not sure if it's really .05 mol/L for Ca(OH)2. I just figured since the molarity was 0.1 N that I could divide 0.1 by 2 hydrogens to give me .05 mol/L

2. I honestly do not know how to even start this..
 
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1.) Prepare 0.2 M solution of Calcium Hydroxide.

2.) Just take 5 gms of NaCl, if we are talking about % w/v, which generally is the case.
 
Why is it 0.2M for #1? And can you show me how to set up for #2 please? Thank you once again!
 
Not 0.2, but 0.05 M Ca(OH)2. 0.05M solution will be 0.05M in Ca2+ and 0.1M in OH- - compare with the definition of normality.

It is not clear if you are to prepare 5% w/w or w/v solution (not that the difference will be large). If w/v - by definition of the w/v it means 5 g in 100 mL of the solution. If w/w - approach depends on how accurate you want to be.

Simple and less accurate approach - dilute solutions have density of 1 g/mL. That means you will need 100 g of solution. Of this 5% is NaCl and 95% is solvent (w/w). Use this information to calculate mass of NaCl.

Exact approach (but most likely overkill) is to check the density of the solution in the density tables (for 5% NaCl it is 1.034 g/mL), calculate mass of the 100 mL - and do exactly the same thing you did in the above case.
 
So for #1 would I set it up like this:

# moles of Ca(OH)2 = 0.100 ml * {0.100 moles of OH(!)/l} * (1 mole of Ca(OH)2 / 2moles of OH)
 
0.100 L, not 0.100 mL. Other than that, looks OK to me.
 

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