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Preparing Solutions and Molarity

  • #1

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..
 

Answers and Replies

  • #2
AGNuke
Gold Member
455
9
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.
 
  • #3
Why is it 0.2M for #1? And can you show me how to set up for #2 please? Thank you once again!
 
  • #4
Borek
Mentor
28,298
2,683
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.
 
  • #5
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)
 
  • #6
Borek
Mentor
28,298
2,683
0.100 L, not 0.100 mL. Other than that, looks OK to me.
 

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