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[Mr. Hyde]

In my Anatomy & Physiology class, our first exam had two Molarity problems on it. I am quite sure I know how to answer molarity problems easily, but these two left me a bit blank on figuring out the percentage of the first problem and the correct solution for the second. Our teacher gave everyone the opportunity to bring our exam questions home and do it at home as well. He is going to average the two exam grades together for the final grade.

The first problem is: 3.3g of calcium chloride was dissolved in 111 ml of water. Calculate the molarity and the percentage of calcium chloride. Calcium chloride, I believe, has an atomic weight of 111. My molarity answer came out as 2.7m. The percentage, I thought, rounded to 30%. However, a part of me is thinking the correct answer is 2.9%. I am not sure what I am doing wrong.

The second problem is: Half of the formula weight (FW) of glucose (in grams) was dissolved in 10 times the formula weight of glucose (in ml). Calculate molarity. Half of the formula weight of glucose, in grams, would be 90. I cannot figure out the other part of the question, so I cannot determine the answer.

Any ideas?

 

[Subliminal]

n(CaCl) = 111/3.3 = 33.6

33.6/111 = 0.30 mol/ml^-3

 

[Mr. Hyde]

n(CaCl) = 111/3.3 = 33.6

33.6/111 = 0.30 mol/ml^-3

Hmm. For the first question, I did it like this.

3.3/111 to get .3 rounded.
I then did 111 x 1000 = 111000.
.3/111000 = 2.7 molarity

I guess I could be completely off base.

 

[Lebron]

Well Molarity is moles of solute over liters of solution. So you did the turning it into moles correctly since it's being dissolved in water. However, you would need to turn 111ml into .111L then divide. Should get like .268 M. Then you would need to turn water into grams using it's denisty, which is 1 g/ml, so it would be 111 grams. Add that to the 3.3, then divide 3.3 by that number times by 100. So it should be 2.88% like you thought.


Also, don't round the number until all your calculations are done and you are doing your sig fig check. It will sometimes throw your numbers way off.

 

[Iceman]

1g/100mL = 1%
100g/100mL = 100%

that's all you really need to know to calculate percentages (or density from a percentage)

so, 3.3g of ANYTHING in 111mLs is 2.97% (roughly 3%).

As for molarity, since 111g in 1L is 1M (1 molar) and your theoretical concentration is 3.3g in 111mL. You convert so your denominator is the same for both, i.e. 111mL x 9.009 = 1L.. and 3.3g x 9.009 = 29.7g.

29.7g (in 1L) is only 26.8% of 111g. That means 3.3g in 111mL is only 0.268 molar (or 268 millimolar).


"Half of the formula weight (FW) of glucose (in grams) was dissolved in 10 times the formula weight of glucose (in ml). Calculate molarity. Half of the formula weight of glucose, in grams, would be 90. I cannot figure out the other part of the question, so I cannot determine the answer."

You'd never encounter a problem like this IRL.

180 g/mol * 0.5 = 90 g/mol.. so 90g of glucose dissolved in 1800 mL (?, why?)

anyway, that's 50g/L and 50g is 0.277 moles of glucose.. that dilution, i.e. 90g glucose in 1.8L of water (?) results in a 0.28 molar solution, or 280millimolar.

 

[Mr. Hyde]

1g/100mL = 1%
100g/100mL = 100%

that's all you really need to know to calculate percentages (or density from a percentage)

so, 3.3g of ANYTHING in 111mLs is 2.97% (roughly 3%).




"Half of the formula weight (FW) of glucose (in grams) was dissolved in 10 times the formula weight of glucose (in ml). Calculate molarity. Half of the formula weight of glucose, in grams, would be 90. I cannot figure out the other part of the question, so I cannot determine the answer."

You'd never encounter a problem like this IRL.

180 g/mol * 0.5 = 90 g/mol.. so 90g of glucose dissolved in 1800 mL (?, why?)

anyway, that's 50g/L and 50g is 0.277 moles of glucose.. that dilution, i.e. 90g glucose in 1.8L of water (?) results in a 0.277 molar solution, or 277millimolar.

Thanks for the help on those. I kept throwing myself off on the second one and that helped make sense of it. I have no idea why he decided to put that question on the exam because none of our molarity problems in class were ever like that.

On the first question, was my solution correct? 2.7m or would it be .268m like Lebron said? I thought I had it right, but now I am questioning myself. I have the percentage thanks to both your help.

 

[Lebron]

Set up the table they taught you. Your units have to cancel out, so that would mean 111mL times 1L/1000mL, which means you divide 111/1000 and L is left since mL have canceled out.

 

[Mr. Hyde]

Set up the table they taught you. Your units have to cancel out, so that would mean 111mL times 1L/1000mL

That's how I ended up with 2.7m.

I took 111 and multiplied it by 1000 (111000). .3/111000 = 2.7027etc

I could be doing something wrong.

 

[Iceman]

On the first question, was my solution correct? 2.7m or would it be .268m like Lebron said? I thought I had it right, but now I am questioning myself. I have the percentage thanks to both your help.

Lebron's right. I just edited in my calculation/logic into my original post.

 

[Mr. Hyde]

Lebron's right. I just edited in my calculation/logic into my original post.

Yeah. I just realized I was doing something completely off base. Thanks to both of you for the help!