q = mcat Chem Problem

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axp107

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I'm getting confused about what to use for the "M" term in q = MCAT

For example:

20 g of of NaCl is poured into a coffee cup calorimeter containing 250ml of water. If the temp. inside drops by 1 degree Celcius by the time the NaCl is totally dissolved, what is the heat of solution for NaCl and water? (Specific heat of water = 4.18 J/g*DegreeCelcius


Upon first thought, I was going to use 20g of NaCl as the Mass. But apparently, you're supposed to use the mass of water (250ml = 250g).

Why?

Ok.. now we continue and plug in etc etc.. we get -1 kJ of heat

Except the answer choices are:

A) -3kj/mol
B)-1kj/mol
C) 1KJ/mol
D) 3 kJ/mol

So I thought... alright.. since we used Water for the mass (for whatever reason) lets divide by the moles of Water. Wrong! We're supposed to divide by the moles of NaCl..

I am sooo confused. I thought q = mcat problems were just plug and chug.. I don't understand the thinking that goes behind them. Can someone explain pls.

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I'm getting confused about what to use for the "M" term in q = MCAT

For example:

20 g of of NaCl is poured into a coffee cup calorimeter containing 250ml of water. If the temp. inside drops by 1 degree Celcius by the time the NaCl is totally dissolved, what is the heat of solution for NaCl and water? (Specific heat of water = 4.18 J/g*DegreeCelcius


Upon first thought, I was going to use 20g of NaCl as the Mass. But apparently, you're supposed to use the mass of water (250ml = 250g).

Why?

For starters they gave you the specific heat of H2O. But water was what you were measuring the temp change of.

Ok.. now we continue and plug in etc etc.. we get -1 kJ of heat

I'm not sure where the /mol part came from here there are no moles in the question stem. Where is this problem? Kaplan? EK? TPR?

Except the answer choices are:

A) -3kj/mol
B)-1kj/mol
C) 1KJ/mol
D) 3 kJ/mol

So I thought... alright.. since we used Water for the mass (for whatever reason) lets divide by the moles of Water. Wrong! We're supposed to divide by the moles of NaCl..

I am sooo confused. I thought q = mcat problems were just plug and chug.. I don't understand the thinking that goes behind them. Can someone explain pls.

TM
 
I'm getting confused about what to use for the "M" term in q = MCAT

For example:

20 g of of NaCl is poured into a coffee cup calorimeter containing 250ml of water. If the temp. inside drops by 1 degree Celcius by the time the NaCl is totally dissolved, what is the heat of solution for NaCl and water? (Specific heat of water = 4.18 J/g*DegreeCelcius


Upon first thought, I was going to use 20g of NaCl as the Mass. But apparently, you're supposed to use the mass of water (250ml = 250g).

Why?

Ok.. now we continue and plug in etc etc.. we get -1 kJ of heat

Except the answer choices are:

A) -3kj/mol
B)-1kj/mol
C) 1KJ/mol
D) 3 kJ/mol

So I thought... alright.. since we used Water for the mass (for whatever reason) lets divide by the moles of Water. Wrong! We're supposed to divide by the moles of NaCl..

I am sooo confused. I thought q = mcat problems were just plug and chug.. I don't understand the thinking that goes behind them. Can someone explain pls.

q = m*c*delta T

Specific heat of water = 4.18 J/g*Degree Celsius. Look at the units for specific heat. Specific means "per unit mass." Since it's the temperature of WATER that's changing, you need to use specific heat of water. So you need to know the MASS of water.

The mass of 250ml of water is 250g since density of water is 1g/1mL.

q = (250g) * (4.18 J/g*Degree Celsius) * (-1 Degree Celsius) = -1000J = -1KJ. So you know
the answer has to be negative, so answer choices C and D are eliminated at this point.

So you are saying this is what they did: NaCl is about 60g/mol, so 20g * 1mol/60g = 1/3 mole. So -1KJ/(1/3 mol) = -3 kJ/mol. So
the answer is A according to the solution?

I agree with you. I don't know why they divide by moles of NaCl and not moles of water. It would help if you posted the problem statement (word for word) and the answer key word for word.
 
see http://www2.wwnorton.com/college/chemistry/gilbert/concepts/chapter13/ch13_1.htm

the heat of solution is, by definition, equal to q divided by moles of solute; therefore, /\H, which is the symbolic representation of the heat of solution, equals "MCAT" divided by the moles of NaCl. Since moles of NaCl equals 1/3 mol, and q=-1kJ, /\H=-1kJ/(1/3)mol=-1*3 kJ/mol =-3kJ/mol, which is answer choice A.
 
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see http://www2.wwnorton.com/college/chemistry/gilbert/concepts/chapter13/ch13_1.htm

the heat of solution is, by definition, equal to q divided by moles of solute; therefore, /\H, which is the symbolic representation of the heat of solution, equals "MCAT" divided by the moles of NaCl. Since moles of NaCl equals 1/3 mol, and q=-1kJ, /\H=-1kJ/(1/3)mol=-1*3 kJ/mol =-3kJ/mol, which is answer choice A.

Yeah, makes sense. Many times things just boil down to a definition.
 
So the "heat of solution" part is just some ambiguous definition then..?

My question really is.. in calorimetry problems, what is the point of the solute then? What does it even do...
If water is being heated.. why are we even given info about the solute?
 
So the "heat of solution" part is just some ambiguous definition then..?

My question really is.. in calorimetry problems, what is the point of the solute then? What does it even do...
If water is being heated.. why are we even given info about the solute?

Water is not being heated. The reaction itself either generates or absorbs heat.
 
I'm getting confused about what to use for the "M" term in q = MCAT

For example:

20 g of of NaCl is poured into a coffee cup calorimeter containing 250ml of water. If the temp. inside drops by 1 degree Celcius by the time the NaCl is totally dissolved, what is the heat of solution for NaCl and water? (Specific heat of water = 4.18 J/g*DegreeCelcius


Upon first thought, I was going to use 20g of NaCl as the Mass. But apparently, you're supposed to use the mass of water (250ml = 250g).

Why?

Ok.. now we continue and plug in etc etc.. we get -1 kJ of heat

Except the answer choices are:

A) -3kj/mol
B)-1kj/mol
C) 1KJ/mol
D) 3 kJ/mol

So I thought... alright.. since we used Water for the mass (for whatever reason) lets divide by the moles of Water. Wrong! We're supposed to divide by the moles of NaCl..

I am sooo confused. I thought q = mcat problems were just plug and chug.. I don't understand the thinking that goes behind them. Can someone explain pls.



Just think of it is that you are trying to determine the the amount of energy transfered to the water. You are putting in the change of temp for the water also so you would never put the mass of the NaCl.
 
There are two ways in which one can report q (heat):

1) As a sum total of the reaction --> Always given in J or kJ --> For the question -1 kJ

Or

2) As per mole of compound placed in calorimeter, kJ/mol --> the question may not indicate which you are looking for but the answer will always be in one or the other!

Hope that helps
 
I'm getting confused about what to use for the "M" term in q = MCAT

For example:

20 g of of NaCl is poured into a coffee cup calorimeter containing 250ml of water. If the temp. inside drops by 1 degree Celcius by the time the NaCl is totally dissolved, what is the heat of solution for NaCl and water? (Specific heat of water = 4.18 J/g*DegreeCelcius


Upon first thought, I was going to use 20g of NaCl as the Mass. But apparently, you're supposed to use the mass of water (250ml = 250g).

Why?

Ok.. now we continue and plug in etc etc.. we get -1 kJ of heat

Except the answer choices are:

A) -3kj/mol
B)-1kj/mol
C) 1KJ/mol
D) 3 kJ/mol

So I thought... alright.. since we used Water for the mass (for whatever reason) lets divide by the moles of Water. Wrong! We're supposed to divide by the moles of NaCl..

I am sooo confused. I thought q = mcat problems were just plug and chug.. I don't understand the thinking that goes behind them. Can someone explain pls.

Let's break this question down to two parts: magnitude and sign. The solution is cooling, so it's an endothermic (heat absorbing) process. This makes deltaH a positive number. We are now down to C versus D.

When a solute is added to solution, the entire solution is heating up. So technically, the mass should be 270 (salt and water are both in solution). Also, technically, the heat capacity of a NaCl solution is higher than pure water, so I'm going to use 4.4 to be anal. We want to determine the amount of heat absorbed. q = mCdeltaT accounts for the solution that undergoes the temperature change. The total heat absorbed by solution is (270 g)(4.4)(1) = 1100 J = 1.1 kJ.

This means that the addition of 20 g NaCl absorbs about 1kJ. Had we instead used a mole, assuming it fully dissociated, then we would have added 60 g instead. This would have absorbed three times as much heat, because there is three times as much dissolving occurring. This means that a mole of NaCl, according to the experiment, would absorb 3 kJ of heat upon full dissociation into water.

The reason the amount of water doesn't matter is because whether we use 250 mL or 500 mL, the same amount of heat will be absorbed upon dissolving 20 g of NaCl.

These questions are in fact plug-n-chug; you just have to follow units when you plug. Brokenglass is again right on the mark with his or her insights.
 
Yup BerkReview is right...

What seems really counterintuitive is that if the temperature of the solution decreases, the reaction is infact ENDOTHERMIC.

Normally .. I just see the words temp decrease and assume exothermic. Grr... how do you pick up on things like this?

I would have picked the negative answer b/c

q = mc (change in T)
where Change in T = T final - T initial

I woulda just plugged in... gotten a negative T since T final dropped... and picked the opposite answer =(
 
Yup BerkReview is right...

What seems really counterintuitive is that if the temperature of the solution decreases, the reaction is infact ENDOTHERMIC.

Grr... how do you pick up on things like this?

In the context of the question, two of the answer choices were Potisitve and two were Negative, so the author of the question wanted you to consider endothermic versus exothermic.

The determination is best done via common sense really. If you touch a flame (an exothermic reaction) it is hot (giving you heat and ultimately burning you). Thus, exothermic processes release heat and raise the surrounding temperature. If you touch an ice cube (an endothermic process) it is cold (absorbing heat from you and ultimately chilling you). Thus, endothermic processes absorb heat and lower the surrounding temperature.

The key thing to keep in mind is: You are the surroundings. If you get cold when touching the system, the system is absorbing your heat, so it is endothermic.
 
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