!Test soon, help please! Chemistry Questions!

[krein123]

Hello! These are my questions:

1) A quantity of ice at 0 degrees Celsius is added to 64.3g of water in a glass at 55 degrees Celsius/ After the ice melted, the temp. of the water in the glass was 15 degrees Celsius. How much ice was added? The heat of fusion of water is 6.01kJ/mol and the specific heat is 4.18J(g x degrees Celsius).

2) The vapor pressure of benzene is 100.00mmHg at 26.1 degrees Celsius and 400.00mmHg at 60.6 degrees Celsius. What is the boiling point of benzene at 760.00mmHg?

Thanks in advance!!

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[Adnama]

Hello! These are my questions:

1) A quantity of ice at 0 degrees Celsius is added to 64.3g of water in a glass at 55 degrees Celsius/ After the ice melted, the temp. of the water in the glass was 15 degrees Celsius. How much ice was added? The heat of fusion of water is 6.01kJ/mol and the specific heat is 4.18J(g x degrees Celsius).

2) The vapor pressure of benzene is 100.00mmHg at 26.1 degrees Celsius and 400.00mmHg at 60.6 degrees Celsius. What is the boiling point of benzene at 760.00mmHg?

Thanks in advance!!

Maybe the answer is.

Q= MCdeltaT
Q=(64.3)(4.18)(40)
Take that Q and use it for the heat of fusion:
Q=mL

 

[Quantum Chem]

I think this might how 1 is done:

Q = mc(delta T)
Q = (64.3g)*(4.18J/Cg)*(15C)
Q = 4031.6 J
4031.6J = Xmoles*Hf
4031.6J = X*(6010J/mol)
.67 = X moles
.67moles*(18g/moles) = 12 grams.

 

[Adnama]

I think this might how 1 is done:

Q = mc(delta T)
Q = (64.3g)*(4.18J/Cg)*(15C)
Q = 4031.6 J
4031.6J = Xmoles*Hf
4031.6J = X*(6010J/mol)
.67 = X moles
.67moles*(18g/moles) = 12 grams.

That makes more sense. My guess was wrong, thanks for clearing it up.

 

[krein123]

thanks so much! have a couple more i am having trouble with....theyre nolality, molarity type problems....

1) An aqueous solution is 8.50% ammonium chloride, NH4CL, by mass. The density of the solution is is 1.024g/mL. What are the molality, mole fraction, and molarity of NH4CL in the solution?

2) A 55-g sample of a gaseous fuel mixture contains 0.51 mole fraction propane, C3H8; the remainder of the mixture is butane, C4H10. What are the masses of propane and butane in the sample?

3) What is the boiling pt. of a solution of 0.150g of glycerol, C3H8O3, in 20.0g of water? What is the freezing point?

4) An aqueous solution of a molecular compound freezes at -0.086 degrees Celsius. What is the molality of the solution?

 

[xIcewind]

Hey krein123 -- I joined to help you, as you joined to ask questions. Haha.

Onto business.

1. You have the percentage composition by mass. Knowing that, for example, 1% by mass of 1L of water (at densilty 1.000g/mL) is 10g, you can calculate the mass of solution.

With the mass you can find thus the mols of NH4Cl in the solution (mols/molar mass/mass relationship). Then, you can apply the definitions of molarity/molality/mole fraction -- molarity is mol/L, molality is mol/kg solvent, and mole fraction is a ratio of two mols.

2) Well. This one you can do sort of algebraically. Assuming there's a mole fraction of 1, for example.

x/(x+y) = 0.51, where x= mol of propane, and y = mol of butane.
x*molar mass of propane + y*molar mass of butane = 55g. Solve!

3) For this you'll need to know the Kf. You can first find the mols of glycerol in 0.150g -- then the molality. Since glycerol has that coefficient thingy of 1 (I apologise, terminology is stuck in my head, the factor where you multiply by the amount of dissociated 'parts' a molecule forms in solution), you can do Tf = Kf * m * i, where i = 1, m = molality, Kf = a constant of temperature (either freezing depression or boiling elevation, but they will be the same absolute number in the end. For example, if the change is 4C, and it normally freezes at 0 and boils at 100, you'll have a final freezing of -4, and a final boiling of 104.)

4) You'll need, again to know the Kf of this compound. Aqueous means that it's in water, so you know that the Tf (from the above Tf = Kf*m*i, and we assume i to be 1, since it's a 'molecular compoud'), we know Kf (or should), and you can find m.

Good luck on your MCATs!

 

[Kaustikos]

That makes more sense. My guess was wrong, thanks for clearing it up.

Actually, you were correct. The delta T = 40, not 15.