Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Wil" on December 29, 2007, 11:53:10 PM
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Please help me solve this question..
A 2.14-L sample of hydrogen chloride(HCl) gas at 2.61 atm and 301K is completely dissolved in 668mL of water to form hydrochloric acid solution. Calculate the molarity of the acid solution. Assume no change in volume.
i have no idea of solving this....
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Remember the ideal gas law? Can you solve for moles?
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Please help me solve this question..
A 2.14-L sample of hydrogen chloride(HCl) gas at 2.61 atm and 301K is completely dissolved in 668mL of water to form hydrochloric acid solution. Calculate the molarity of the acid solution. Assume no change in volume.
i have no idea of solving this....
You might become more familiar with what to do if you convert your units for use with the ideal gas law. :)
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PV = nRT...and remeber to check you units
rearrange this and
n=PV/RT
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I'm not familiar with molarity, could someone explain this to me? Thanks!
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PV = nRT is the ideal gas law. This law states that a 1.o mol sample of gas will occupy 22.4L at STP. Where STP can have different units. My favorite are R = 0.08205784 L·atm·K-1·mol-1.
Molarity means moles/L it is a unit of concentration....it plays into this problem beacause all acids are gases in thier natural elemental state. They are made into liquids because the gases are dissolved into water.
Example: HCl is hydrochloric acid and is naturally a gas. When it is dissolved in water it becomes a liquid. If you ever read the label on a bottle of acid in a lab this shows you that no acid (liquid form is 100%).
So you have to work with molarity. A mole of HCl = gfw = 36.5 g/mole if you want to prepare a 1M solution you would weigh out 36.5 g and place it in a 1L flask and dilute to 1L. That would be your 1M HCl solution...
here you have n =PV/RT
you are given:
V = 2.14L HCl
P = 2.61 atm
T = 301K
Let R = 0.0821 L·atm·K-1·mol-1 (so your units cancel)
and the 2.14L of HCl is dissolved in 668 mL ( 0.668L)of water to make an HCl acid solution
n =PV/RT....you need to find your moles of gas first
then express them as a concentration by dividing by the volume of water they are dissolved in
Plug the listed variables into the n = PV/RT and solve for n
take n and divide by 0.668 L (have to keep units straight convert the 668 mL to L)
Then you have mol/L which is M which is molariy...
I will check your answer
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So how would you solve the probem below where you have to find the substance that is placed into the water.
"Something which would lower the freezing point of water to -3.72 degrees Centigrade if one mole is placed in 1.00 Kg of H2O"
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The answer is a salt....and the process is known as freezing point depression...
Water becomes ice at a temperature of 32 degrees Fahrenheit (0 degrees Celsius), which we call the freezing point of water. Adding salt to water lowers the freezing point, thus preventing the formation of ice.
The primary difference between water and ice is the speed at which the molecules move. In water, the molecules move more rapidly than in ice. The temperature of the water is changed by adding or removing heat. As heat is removed, the water molecules slow down. At the freezing point, the motion of the molecules is slow enough that the water becomes ice.
When salt is added to water, the salt and water molecules stick together, <ore heat then must be removed to slow down and separate the combined salt and water molecules. The removal of the additional heat is required to freeze the salt-water combination. This means that the temperature at which it freezes is below that of pure water.
Antifreeze uses the same principle to keep the water in your car from freezing in the winter. Spreading salt over roads prevents melting snow from freezing into ice and causing hazardous road conditions.
For similar reasons, the addition of salt to water increases the boiling point of the water: more heat must be added to make the larger combined salt plus water molecule move fast enough to boil than must be added to make plain water boil. The result is that the temperature of the boiling salted water is higher than that of plain water.
Salt is not the only substance which lowers the freezing point of water. In fact, anything that can be dissolved in water will have the same effect.
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Molarity means moles/L it is a unit of concentration....it plays into this problem beacause all acids are gases in thier natural elemental state. They are made into liquids because the gases are dissolved into water.
Beware when stating "all". What about liquid H2SO4? Solid oxalic?
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So any salt, whether its NaCl or CaCl2, would fit in that equation? Different salts don't lower the freezing point to different temperatures?
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So any salt, whether its NaCl or CaCl2, would fit in that equation? Different salts don't lower the freezing point to different temperatures?
http://en.wikipedia.org/wiki/Freezing-point_depression
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So any salt, whether its NaCl or CaCl2, would fit in that equation? Different salts don't lower the freezing point to different temperatures?
According to the ideal Colligative properties, each solute lowers vapor pressure raises boiling point, lowers freezing point, and increases osmotic pressure. So a non-ionic solutes, like ethylene glycol or sucrose, count as one, NaCl counts as two solutes (Na+ and Cl-) CaCl2 counts as 3, Ammonium phosphate counts as 4, etc. However, ideal is just that, a theoretical construct. In real life chemicals don't behave perfectly ideally.
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Sure...specifically, HCl is a gas until dissolved in water. Can deal with H2SO4 and Oxalic (great chelation agent) at another time.
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What is the formula for the universal gas constant when solving for the cryoscopic constant? I am confused with what goes on the top and what goes on the bottom of the fraction.
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Would I be right to say that heat of fusion goes on top while Kelvins and number of moles goes on the bottom? Or do i have the wrong units all together?
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For what? I need a question to give you an answer!!!! Or am I missing something?
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Maybe this will help you...is this the cryoscpic question...all about molality...P-Chem see Atkins text:
When working with boiling point elevations or freezing point depressions of solutions, it is convenient to express the solute concentration in terms of its molality m defined by the relation:
For this unit of concentration, the boiling point elevation, Tb - Tºb or DTb, and the freezing point depression, Tºf - Tf or DTf, in ºC at low concentrations are given by the equations:
DTb = kbm DTf = kfm
where kb and kf are characteristic of the solvent used. For water, kb = 0.52 and kf = 1.86. For benzene, kb = 2.53 and kf = 5.10.
One of the main uses of the colligative properties of solutions is in connection with the determination of the molar masses of unknown substances. If we dissolve a known amount of solute in a given amount of solvent and measure DTb or DTf of the solution produced, and if we know the appropriate k for the solvent, we can find the molality and hence the molar mass, MM, of the solute. In the case of the freezing point depression, the reaction would be:
See attachment