hello everyone, i have been working o n these 3 problems since monday and i have yet to get the rigth answers. I am getting very frustrated.. and i need every point i can get in this class just in case i do bad on an exam. please help, with these questions. Of course the book examples are easy.. but the homework ones seem insane.

1. You are asked to prepare an aqueous solution of ethylene glycol (HOCH2CH2OH) with a mole fraction of 0.199.

If you use 569 g of water, what mass (in g) of ethylene glycol should you use?

What is the molality of the resulting solution?

2. At 25oC, the equilibrium pressure of HCl vapor above a 0.100 M aqueous solution of HCl is 4.0 torr.

Calculate the Henery's law constant.

What is the equilibrium pressure (in torr) of HCl vapor above a 6.81 M solution?

3. A solution contains 11.00 g of sucrose (C12H22O11) in 161 mL of water. Assume the density of water is 1.00 g/mL.

Calculate the freezing point of the solution. The normal freezing point of water is 0.00 oC. The Kf of water is 1.858 oC kg/mol.

Calculate the boiling point of the solution. The normal freezing point of water is 100.00 oC. The Kb of water is 0.512 oC kg/mol.

just to help you begin with....

Molality of a solution may be defined as the number of gram mole of the solute disolved in 1000 gm ( 1Kg) of the solvent.

if we go to the equation then this is what we get.

mathematically, Molality = Gram mole of solute

_____________________________

Mass of the solvent ( Kg)

Mass of solute ( g)

or , = _____________________________________________

(GMM of solute) X ( Mass of solvent (Kg))

i hope that should help.

1. You are asked to prepare an aqueous solution of ethylene glycol (HOCH2CH2OH) with a mole fraction of 0.199. If you use 569 g of water, what mass (in g) of ethylene glycol should you use?

What is the molality of the resulting solution?

ethylene glycol = eg

mole fraction of eg = moles of eg/ (moles of eg + moles of water)

We are given the mass of water so we can find the number of moles of water by dividing by the mole mass of water. Then plug into the mole fraction equation and solve for moles of eg.

Once we know the moles of eg, we can plug into the molality formula

molality = moles of eg/ kg of water (make sure to convert from grams to kg)

2. At 25oC, the equilibrium pressure of HCl vapor above a 0.100 M aqueous solution of HCl is 4.0 torr.

Calculate the Henery's law constant.

What is the equilibrium pressure (in torr) of HCl vapor above a 6.81 M solution?

Henry's Law states

[HCL ] = k x p

_{HCl}Solving for the constant yields:

k = [HCl]/ p

_{HCl}= .100 M / 4.00 Torr = .025 M/TorUsing the value for k above along with the p

_{HCl}= 6.81 Torr[HCl] = .025 M/Torr x 6.81 Torr

3. A solution contains 11.00 g of sucrose (C12H22O11) in 161 mL of water. Assume the density of water is 1.00 g/mL.

Calculate the freezing point of the solution. The normal freezing point of water is 0.00 oC. The Kf of water is 1.858 oC kg/mol.

Calculate the boiling point of the solution. The normal freezing point of water is 100.00 oC. The Kb of water is 0.512 oC kg/mol.

Use the density of water to find the mass of the water (the solvent).

Find the number of moles of sucrose (the solute) by dividing by the molecular mass of sucrose.

Calculate the molality of the solution by dividing the moles of solute by the mass of the solvent (IN KILOGRAMS).

Change in FPt = Kf x molality

Change in BPt = Kb x molality

thanks for the hepl everyone.

I managed to get one answer right. The other 2, I am still having trouble with. Here is what I have.

1. You are asked to prepare an aqueous solution of ethylene glycol (HOCH2CH2OH) with a mole fraction of 0.199.

If you use 569 g of water, what mass (in g) of ethylene glycol should you use?

What is the molality of the resulting solution?

answer: (mol solute + mol solvent = mol solution), so we take: (mol solute/ mol solution)

Since 569 g of water is being used, we divide it by the molar mass of H2O, which equals (18 g /mol):

(569gH20 x mol H20/18 g H20 = 31.6 mol)

Ethylene glycol has a mole fraction of 0.199, so remembering the fractions on the top:

.199 = (mol solute / mol solute + 31.6 g solvent)

rearrange: 0.199 (31.6 + mol solute ) = mol solute

6.3 + 1 = mol solute

7.3 = mol solute

The molar mass for (HOCH2CH2OH) is 62 g/mol, so:

(62 g /mol x 7.3 mol) = 452.6 g Ethylene glycol

2. 2. At 25oC, the equilibrium pressure of HCl vapor above a 0.100 M aqueous solution of HCl is 4.0 torr.

Calculate the Henery's law constant.

What is the equilibrium pressure (in torr) of HCl vapor above a 6.81 M solution?

The solubility of HCl = (Henry’s law’s constant) x (Partial pressure)

HCl = kH x pHCl

Rearrange and solve for the constant: kH = (HCl)/ pHCl

kH = (.100 M ) / (4.00 torr)

kH = .025 M / torr

Finally, using the original formula for Henry’s law constant: (HCl = kH x pHCl)

= 6.81 M x .25 M/torr

= 1.7025 torr

.199 = (mol solute / mol solute + 31.6 g solvent)

Your problem is in the next statements

rearrange: 0.199 (31.6 + mol solute ) = mol solute

6.3 + 1= mol solute

You made a math error here, let x = mol solute and plug that into the equation before the one in red:

0.199 ( 31.6 + x) = x

Multiplying we get

6.33 + .199x = x

Solving for x by subtracting .199x from both sides we get

6.3 = .801x

x = 6.3 /.801 = 7.8 mol solute

Now re-calculate the mass of ethylene glycol and it should be correct.

Problem 2:

Finally, using the original formula for Henry’s law constant: (HCl = kH x pHCl)

You made your mistake when you plugged your numbers in

= 6.81 M x .25 M/torr = 1.7025 torr

It should be 6.81 M = .25 M/torr x pHCl

pHCl = 6.81 M / .25 M/torr = 27.24 Torr

i'm still not coming up with the right answer at all, i have to submit these answers online so even if i am close it wont accept it. it has to be exact.

the 2 questions i posted are making me insane, by any chance would anyone be willing to walk me step through step so i can get the right answer?

In question 1, the correct answer should be the product of the number of moles of ethylene glycol times the molar mass.

7.8 moles x 62 g/mol = 483.6 g of ethylene glycol

The way that you had it set up was fine with the exception of the math error that I highlighted in red in the previous post

In question 2, your set-up was again fine with the exception of how you plugged in. You substituted in the Molarity in place of the pressure. I corrected that by plugging into the equation correctly. The equation with answer is shown again below:

It should be 6.81 M = .25 M/torr x pHCl

pHCl = 6.81 M / .25 M/torr = 27.24 Torr

last question.. thanks everyone for all the help.

this question is for extra credit.

Ethylene glycol, an automotive coolant, has the chemical formula HOCH2CH2OH. Calculate the vapor pressure (in torr) above a coolant solution containing 57.3 g of ethylene glycol and 0.751 L of water (density = 1.00 g/mL) at 100oC, the normal boiling point of pure water.

how would i go about answewring this.

You want to find the vapor pressure of a solution. Raoult's Law would seem to be the best way of doing that.

Psolution = p pure solvent x mole fraction of solvent.

The vapor pressure of pure water at 100C is 760 Torr (at 100C pure water boils because the vapor pressure of the water equals the normal atmospheric pressure of 760 Torr.)

To find the mole fraction of the water, we need to find the moles of e.g. present as well as the moles of water.

Once you know the moles of solute and the moles of solvent, you can calculate the mole fraction of the water (the solvent)

mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)

Plug into Raoult's law and find the vapor pressure above the solution.

hey.. I need help solving these 3 problems, I have been working on them for quite sometime and am having trouble coming up with the correct answere. If any one can help me, it will be greatly appreciated. Thank You

1. When camping in remote locations, water can be in short supply. It is not

practical to carry all of the water needed during a trip from the starting

point to the campsite. Hikers often look for other sources of potable

(drinkable) water near their campsite. During the winter months, it is

possible to melt clean snow and turn it into drinkable water. To warm

themselves, campers make pots of “trail coffee” – coffee grounds boiled in

a pot with water.

a. Two hikers in Acadia National Park melt snow to make 1000. mL of

trail coffee. If the snow used is originally at –10 °C, how much

energy is required to heat the water to its boiling point? [Cp(ice) =

37.1 J/mol °C; ?Hfus = 6.01 kJ/mol; Cp(H2O) = 75.3 J/mol °C]

Assume that the density of snow and liquid water are both 1.00

g/mL.

b. To boil the water, a small propane-fueled stove is commonly used.

Given that the heat of combustion (?H°c) of propane (C3H8) is –

2220. kJ mol-1, please calculate the number of moles of propane

needed to melt the water.

c. Propane is usually sold in 400 g fuel canisters. How many canisters

would you need to make the coffee?

2. A 1.00% by mass NaCl(aq) solution has a freezing point of -0.192 C.

a. Estimate the van’t Hoff factor (i) from the data.

b. Determine the molal concentrations of Na+. Cl-, and NaCl.

c. Calculate the percent dissociation of NaCl in this solution.

(Hint: When calculating freezing point depression, the

concentration (m*i) is the sum of the molalities of Na+. Cl-, and

undissociated NaCl.)

3. In the human liver, the oxidation of ethanol (from alcoholic beverages) to

acetaldehyde, catalyzed by the enzyme by alcohol dehydrogenase, is a

zeroth-order reaction.

CH3CH2OH + NAD+ ? CH3CHO + NADH + H+

An average 70-kg person typically takes about 2.5 hrs to oxidize the 15 mL

of ethanol (C2H5OH; density = 0.7893 g mL-1) contained in a single 12-oz

can of beer, 5-oz glass of wine, or one shot of distilled spirits.

a. Which plot will be linear - [ethanol] vs. time, ln [ethanol] vs. time,

or 1/[ethanol] vs. time?

b. What is the rate law for this reaction?

c. What is the value of the rate constant for this reaction? Be sure to

include the proper units.

d. The actual rate of oxidation varies greatly from person to person.

What causes this discrepancy? (Assume that all people have the

same body temperature of 98.6 ºF.)