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Topic: uh i actually have no idea? solutions?  (Read 26336 times)

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Offline BAMxTiFFANY

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uh i actually have no idea? solutions?
« on: March 15, 2009, 03:32:53 PM »
so i have this mini quiz thing and im really confused
i need to show work i've gotten 2 of 7
so if you can show me how you did it it would be great
we need to assume like 1 liter of solution or 1 kg of it x]


1. i did it
2. a sodium chloride solution has a molarity of 1.250 mol/L and a density of 1.050g/ml. calculate molality, the mole fraction of water, and the percent by mass of sodium chloride.
3. a solution of potassium nitrated has a molality of .00350 mol/kg and a denisty of 1.009g/ml. calculate the mole fraction of potassium nitrate and the molarity.
    ^ i got the molarity already
4. one gram of sodium nitrate was dissolved in 100.0 ml of water. the new volume of this solution is 100.1 ml. the density of water is 1.00 g/ml. calculate the molarity, molality, mole fraction of water and the density of this new solution
5.the molarity and molality of a potassium chloride solution are 1.00mol/l and .9750 mol/kg,respectively. calculate the solution's density.
6. ten (10.00) grams of sodium chloride are dissolved in enough water to give you 100.0 ml of solution.  one millilter of this new solution is added to enough water to give you a new volume of 110.0 ml. what is the molarity of this new solution?
7. i got it


i would love it if someone helped x]

Offline ARGOS++

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Re: uh i actually have no idea? solutions?
« Reply #1 on: March 15, 2009, 04:03:30 PM »

Dear BAMxTiFFANY;

Can you show us your starting work for each question you ask that we can guide you?

Do you know the definition of molarity, molality, and mole fraction?   –   Otherwise you may start to ask Wiki for them.

Good Luck!
                    ARGOS++

Offline BAMxTiFFANY

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Re: uh i actually have no idea? solutions?
« Reply #2 on: March 15, 2009, 04:18:36 PM »

Dear BAMxTiFFANY;

Can you show us your starting work for each question you ask that we can guide you?

Do you know the definition of molarity, molality, and mole fraction?   –   Otherwise you may start to ask Wiki for them.

Good Luck!
                    ARGOS++


yup i do x]

mole fraction > mole of substance over total moles
molarity > mole of solute over liters over solvent
molality > mole of solute over kg of solvent

Offline ARGOS++

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Re: uh i actually have no idea? solutions?
« Reply #3 on: March 15, 2009, 04:46:27 PM »

Dear BAMxTiFFANY;

Except molarity all is correct!
Please try this one again and then start answering question 2.

Good Luck!
                    ARGOS++

Offline BAMxTiFFANY

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Re: uh i actually have no idea? solutions?
« Reply #4 on: March 15, 2009, 05:12:29 PM »

Dear BAMxTiFFANY;

Except molarity all is correct!
Please try this one again and then start answering question 2.

Good Luck!
                    ARGOS++


The number of moles of solute per liter of solution. <molarity O_o my bad
but wut does this have to do?
i dnt how to start this..

Offline ARGOS++

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Re: uh i actually have no idea? solutions?
« Reply #5 on: March 15, 2009, 05:21:45 PM »

Dear BAMxTiFFANY;

Now it’s also ok!

For 2: Can you calculate for the begin who many grams sodium chloride are in 1.00 L solution dissolved, using your definition of molarity?

Good Luck!
                    ARGOS++

Offline BAMxTiFFANY

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Re: uh i actually have no idea? solutions?
« Reply #6 on: March 15, 2009, 05:43:06 PM »

Dear BAMxTiFFANY;

Now it’s also ok!

For 2: Can you calculate for the begin who many grams sodium chloride are in 1.00 L solution dissolved, using your definition of molarity?

Good Luck!
                    ARGOS++

huh O_o begin? who?
like..?


1.050g = xg
1 ml       1000mL (1 L)?   so x is.. 1050g dissolved in 1 liter


is that right?

Offline ARGOS++

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Re: uh i actually have no idea? solutions?
« Reply #7 on: March 15, 2009, 05:50:58 PM »

Dear BAMxTiFFANY;

No!, - that’s not correct.

Let’s try it simpler:
   a.)   How much grams is 1.25 mole of sodium chloride?
   b.)   And then think about that so much is dissolved in 1.00 L solution!

Good Luck!
                    ARGOS++

Offline BAMxTiFFANY

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Re: uh i actually have no idea? solutions?
« Reply #8 on: March 15, 2009, 05:55:44 PM »

Dear BAMxTiFFANY;

No!, - that’s not correct.

Let’s try it simpler:
   a.)   How much grams is 1.25 mole of sodium chloride?
   b.)   And then think about that so much is dissolved in 1.00 L solution!

Good Luck!
                    ARGOS++

a) 73.05
b) is it at STP O_o

Offline ARGOS++

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Re: uh i actually have no idea? solutions?
« Reply #9 on: March 15, 2009, 06:02:44 PM »

Dear BAMxTiFFANY;

That’s exact!  -  As you are not dealing with gas you must not care about STP.

Now for the next step:
     -   And now can you tell what the weight of this 1.00 L solution is by using the density?
     -   How many grams water is contained in this 1.00 L solution?
     -   And with all that you are able to calculate the molality of the solution using the definition of.

Good Luck!
                    ARGOS++

Offline BAMxTiFFANY

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Re: uh i actually have no idea? solutions?
« Reply #10 on: March 15, 2009, 06:09:47 PM »

Dear BAMxTiFFANY;

That’s exact!  -  As you are not dealing with gas you must not care about STP.

Now for the next step:
     -   And now can you tell what the weight of this 1.00 L solution is by using the density?
     -   How many grams water is contained in this 1.00 L solution?
     -   And with all that you are able to calculate the molality of the solution using the definition of.

Good Luck!
                    ARGOS++

1. 73.05g/L
2. .804 g?

2. 

Offline ARGOS++

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Re: uh i actually have no idea? solutions?
« Reply #11 on: March 15, 2009, 06:19:30 PM »

Dear BAMxTiFFANY;

Your 1.) Is correct for the concentration, but not for the whole weight of the solution.
Only after your 1.) is correct, you are able to calculate how much water is in this 1.00 L solution.
And finally you can calculate the molality.

So please try once again.
Good Luck!
                    ARGOS++

Offline BAMxTiFFANY

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Re: uh i actually have no idea? solutions?
« Reply #12 on: March 15, 2009, 06:27:47 PM »

Dear BAMxTiFFANY;

Your 1.) Is correct for the concentration, but not for the whole weight of the solution.
Only after your 1.) is correct, you are able to calculate how much water is in this 1.00 L solution.
And finally you can calculate the molality.

So please try once again.
Good Luck!
                    ARGOS++


you just confused me -_-
ughh kill me .__.

Offline ARGOS++

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Re: uh i actually have no idea? solutions?
« Reply #13 on: March 15, 2009, 06:31:47 PM »

Dear BAMxTiFFANY;

Let’s try it as simple as before:
    If you know the density of 1.050 g/ml, then what is the weight of 1.00 L solution?

Good Luck!
                    ARGOS++

Offline BAMxTiFFANY

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Re: uh i actually have no idea? solutions?
« Reply #14 on: March 15, 2009, 06:35:23 PM »

Dear BAMxTiFFANY;

Let’s try it as simple as before:
    If you know the density of 1.050 g/ml, then what is the weight of 1.00 L solution?

Good Luck!
                    ARGOS++


uh O_o
 g/v so.. O_O 1050 ? -_-
ugh do you have aim?

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