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Molarity problem

  • Thread starter ~angel~
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  • #1
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I just wanted to see if I'm doing this right.

Sulfuric acid is supplied as a concentrated liquid. A bottle was assayed and found to have a density of 1.84 g/mL and a purity of 97%. The molecular weight of H2SO4 is 98. What is the molarity of this solution?

molarity=moles/volume
moles=mass/mol.weight
moles=1.84/98
moles=1.88*10^-2

assuming the volume is 1mL=0.001L,

molarity=1.88*10^-2/0.001
molarity=18.80 mol/L

since there is 97% purity,

molarity=18.80*0.97
=18.24 mol/L

Is this right? I think I'm wrong, but this was just a guess.

Thanks.
 

Answers and Replies

  • #2
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I guess so.
 
  • #3
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Thanks.

I just have one more question.

A solution requires 250mL of a 7.5% tri sodium citrate. The only cirate you have available is the dihydrate C6H5O7Na3.2H2O with a formula weight of 294.1. How much would you have to weigh out to make up this solution?

I'm not sure how to do this, so could some one show me how to do it please?

Thanks.
 
  • #4
saltydog
Science Advisor
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~angel~ said:
I just wanted to see if I'm doing this right.

Sulfuric acid is supplied as a concentrated liquid. A bottle was assayed and found to have a density of 1.84 g/mL and a purity of 97%. The molecular weight of H2SO4 is 98. What is the molarity of this solution?

molarity=moles/volume
moles=mass/mol.weight
moles=1.84/98
moles=1.88*10^-2

assuming the volume is 1mL=0.001L,

molarity=1.88*10^-2/0.001
molarity=18.80 mol/L

since there is 97% purity,

molarity=18.80*0.97
=18.24 mol/L

Is this right? I think I'm wrong, but this was just a guess.

Thanks.
Well, when you say 97% pure, I'll assume you me by weight. From the density, the weight of 1 liter is 1840 grams. Now, 97% of that would be 1784.8 grams [itex]H_2SO_4[/itex]. Since every mole of the acid has 98 grams, thus in the 1 liter bottle, we'd have 1784.8/98 moles of the acid or the concentration would be 18.21 M. What is the standard molar concentration of the acid you'd find on the bench in a college Chem lab? Is it 18.21?
 
  • #5
saltydog
Science Advisor
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~angel~ said:
Thanks.

I just have one more question.

A solution requires 250mL of a 7.5% tri sodium citrate. The only cirate you have available is the dihydrate C6H5O7Na3.2H2O with a formula weight of 294.1. How much would you have to weigh out to make up this solution?

I'm not sure how to do this, so could some one show me how to do it please?

Thanks.
Well, how about this: Make 1 kg of solution. That means you'd need 75 grams of the anhydrous sodium citrate right? Well, surely you'll need more than 75 grams of the hydrate right? What is the ratio of hydrate/anhydrous in the dihydrate? So if I needed 1 gram of anhydrate, I need 1 times that ratio of the hydrate. Add it in a beaker, add enough water to make 1000 grams total, take 250.
 
  • #6
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Thank you so much salty dog :)

Could you also help lead me in the right direction for this question if you don't mind.

One of the components of the mediums in which worms are grown is called S basal. It is composed of 0.1M NaCl, 0.05M potassium phosphate and 0.0005% (w/v) cholesterol. You have the foloowing stock solutions and powders available: NaCl powder (mol. wt. 58.4), 1M potassium phosphate, and a 5mg/mL stock of cholesterol (mol wt. 386.7). How would you make up 250mL of S basal?

Thank you. :)
 
  • #7
saltydog
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~angel~ said:
Thank you so much salty dog :)

Could you also help lead me in the right direction for this question if you don't mind.

One of the components of the mediums in which worms are grown is called S basal. It is composed of 0.1M NaCl, 0.05M potassium phosphate and 0.0005% (w/v) cholesterol. You have the foloowing stock solutions and powders available: NaCl powder (mol. wt. 58.4), 1M potassium phosphate, and a 5mg/mL stock of cholesterol (mol wt. 386.7). How would you make up 250mL of S basal?

Thank you. :)
Well again, make up 1 liter of solution but add the water up to 1 liter last right. Add the right amount of NaCl and KPO4 and cholesterol into a beaker, then fill it up to 1 liter. Start with the cholesterol. If it's 0.0005 volume % then doesn't this relation hold:

[tex]0.0005=\frac{x}{1000 ml} 100[/tex]

So figure out how much of the stock cholesterol solution you need for that amount of grams and add it first to the beaker.

The salt is easy right. You want a 0.1 M solution and you're making up a 1 liter solution so how much salt you need? Add it to the pot.

Now, you want a 0.05M solution of KPO4 but have a 1 M solution. Well, if I took 100 ml of that 1 M solution and diluted it to 1000 ml, wouldn't that be a 0.1 M solution? How many ml of the 1 M solution then would I have to take (and dilute to 1000 mL) to get a 0.05 M? Add it to the pot (beaker).

Now, finally fill up the beaker to 1000 ml with water.

I think this is right. You'll need to use your own judgement and go over it and make sure you understand what's going on in case I made an error.
 

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