Perform the following conversions

Question

A.) 50.0mL NH₃(g) at -20°C and in 75.0 ml of solution to concentration (molarity)

B.) Volume (in liters) of a 0.250 M solution that would release 5.25 L of SO₃ (g) at STP

C.) Volume (in liters) of 2.65 M solution that would release 40.0 L of NO₂(g) at 75°C and 500 torr

 

M=molarity=concentration=(mol/L)

Answer 1

A)50.0mL NH3(g) at -20°C and in 75.0 ml of solution to concentration (molarity

Convert 50 ml NH3 gas at -20ºC to moles: In order to do this, one really needs to know the pressure, but assuming it to be 1 atm, then from PV = nRT and solving for n, one gets... n = PV/RT = (1atm)(0.05L)/(0.0821L-atm/deg-mole)(253deg), and n = 0.0024 moles of NH3.  NOTE: 50 ml becomes 0.05 L and -20ºC becomes 253ºK.

molarity = moles/liter = 0.0024 moles/0.075L = 0.032 M

 

B)Volume (in liters) of a 0.250 M solution that would release 5.25 L of SO3 (g) at STP

5.25 L SO3(g) at STP is how many moles?  1 mole = 22.4 L, thus 5.25 L/22.4 L/mol = 0.234 moles of SO3 g

From a solution that is 0.250 M, that would be 0.25 moles/L, and 0.25 moles/L (x L) = 0.234 moles

Solving for X, one gets 0.938 liters

 

C)Volume (in liters) of 2.65 M solution that would release 40.0 L of NO2(g) at 75°C and 500 torr

40.0 L NO2(g) is how many moles under the given conditions?  PV = nRT and n = PV/RT

Converting 500 torr to atm, one get 500 torr x 1 atm/760 torr = 0.6579 atm 

n = (0.6579 atm)(40 L)/(0.0821 L-atm/deg-mol)(348 deg) = 0.921 moles NO2 gas

From the solution that is 2.65 mol/L, how many liters are needed to obtain 0.000921 moles?

2.65 mol/L (X L) = 0.921 moles and 

X = 0.348 Liters 

NOTE:  you can leave the pressure in torr (500 torr) but then you must use a different R (gas constant) value. That value would be 62.36 L-torr/mol-deg

 

PS. I can be available for online tutoring.

Answer 2

50 ml = 0.05 L; but, oops, I see -20 C.  So first we must use Charles' Law to find the volume at -20C:

 

V = T(V1/T1) = 253 K (0.05/273K) = 0.046336996336996336996336996337 Liters of gas.  (That makes sense - the colder the less the volume.) Also sorry about all the numbers, but I love the Window's calculator and memory.  Using 3 sig.figs., it rounds to 0.0463.

 

0.0463 L of NH₃ is how many moles of NH₃ ?

 

One mole of any gas occupies 22.4 Liters, so again how many moles is 0.0463 L of any gas ?

 

It is a fraction of a mole.  What fraction?

 

         0.0463 L

It is  --------------   =    0.00206861590790162218733647305076 moles

        22.4L/mole

 

(again, sorry, but I love the computer calculator and memory, and I really don't like to round off until the end.)

 

So we have 0.00223214285714285714285714285714 moles of NH_3, and add a sufficient quantity of water to make 75 ml.

 

O, oh.  What level chemistry is this?  As a weak acid, we could be dealing with the weak acid dissociation and the dissociation of water:

 

NH₃ +  HOH  <--->  NH₄⁺   +    OH^-, and then dealing with the dissociation constants K_b and maybe K_w .

 

Well, I am going to ignore it, because maybe I'd be doing a lot more work than I have time for right now.  It would make this problem more of a high school project than a practice problem.

 

Molarity = moles/liter

 

M          = 0.00223 moles / 0.075 L

 

M          =  0.02758154543868829583115297401012 molar NH₃

 

 

It is about 0.028 M NH₃ .  (I think. Maybe a bit lower since a small amount will react with water.)

 

I am not going to do the rest.  Sorry, but I just jumped into this and I haven't done Chemistry for a long time so I thought it was going to be easier and teachable.

 

I have already spent a long time on this so I won't trash it.  
Well, you might learn something from it.

If this is a project, to teach it would take a lot more than what I just did.  Each part of the project would need practice until you know the pieces well enough to fit them together into the project.