Water questions (1 Viewer)

[Dragonmaster262]

Q) Calculate the molar heat of the solution for ionic substances involved in the experiment which was carried out in well insulated plastic beakers using a thermometer which read to 0.1 C. Take the specific heat capacity of the fnal solutions as 4.2 J/k/g. 5.3 g calcium chloride was dissolved in 250 ml water at 18.6 g. The temperature rose to 22.0 C.

 

[Dragonmaster262]

Any assistance please?

 

[Gussy Booo]

This is hard... ummm

I could help...i think..........................................you're doing this in prelim? wow....
I think we're on the same topics in chemistry, but i started Energy the other day. anyways.

This looks like it has to do with Q=mC[DELTA T].

So, the first thing I can see from your statement is the change in temperature.

so thats. (22-0.1)

So T = 21.9 ......
C is given to us....4.2

I'm not to sure what the mass might be...

It should be 250...

What do you think?

I get lost at : 5.3 g calcium chloride was dissolved in 250 ml water at 18.6 g

 

[Dragonmaster262]

This is hard... ummm

I could help...i think..........................................you're doing this in prelim? wow....
I think we're on the same topics in chemistry, but i started Energy the other day. anyways.

This looks like it has to do with Q=mC[DELTA T].

So, the first thing I can see from your statement is the change in temperature.

so thats. (22-0.1)

So T = 21.9 ......
C is given to us....4.2

I'm not to sure what the mass might be...

It should be 250...

What do you think?

I get lost at : 5.3 g calcium chloride was dissolved in 250 ml water at 18.6 g

You're right it is 250 g.

 

[Dragonmaster262]

If you need the specific question go to Preliminary Conquering Chemistry page 241, question 34.

 

[Gussy Booo]

Dam, I use Chemistry Contexts.

 

[HaydenLee]

If you need the specific question go to Preliminary Conquering Chemistry page 241, question 34.

It's actually page 226, in 4th edition. note that it is 34 a) and that the question says the water is at 18.6°C, not 18.6 g. The question as written in my book is:

34 Calculate the molar heat of solution for the ionic substance involved in each of the following experiments which were carried out in well insulated plastic beakers using athermometer which read to 0.1°C. Take the specific heat capacity of the final solutions as 4.2 J/K/g
a) 5.3 g calcium chloride was dissolved in 250 mL (=250 g) water at 18.6°C. The temperature rose to 22.0°C

Hope that clarifies the question; it should now be easier to answer.
Cheers,
HJL

 

[Gussy Booo]

LMFAOO!! THAT MAKES THINGS A WHOLE LOT EASIERR !!! ILL CHANGE MY THING AND DO THE ANSWERS AGAIN.


And no, we're not using Q = mC[Delta T], we're going to use [DELTA H] = -mC[DeltaT].

We're using this, because the question concerns exothermic/endothermic procedures.

So like, I said, there are two steps.

STEP 1

So we apply [DELTA H] = -mC[DeltaT].

M = 250 [Given by water]
C = 4.2 [Just to let you know, it should be 4.18, because that's the real heat capacity of water, but the question has rounded it up to 4.2]

Delta T = (22.0-18.6)
Delta T = 3.4

So lets put these 3 things into the formula

DELTA H = - ( 250 x 4.2 x 3.4 )
DELTA H = -3570 J per gram.

That's step one......

STEP 2

Now we have to find the Molecular Mass of Calcium Chloride.

Ca = 40.08
Cl = 35.45

M(CaCl) = 75.53g per mole.

Now we take our last result and times them together.

So...

(-3570 x 75.53)

= -269.64 Kj per Mole........orr.... -269642.1 J per mole..


BUTT I THINK IM WRONG..BECAUSE IM NOT TAKING THE 5.3 G OF CALCIUM CHLORIDE intoo consideration when calculating that ! ughh ive forgotten all of this during the holidays !

its soo frustrating !

 

[HaydenLee]

So close, Gussy!
You have mostly the right process.
Firstly, Calcium has a valency of -2, so the formula is CaCl2. Therefore the number of moles (what you actually want to find) of calcium chloride is 5.3/(40.08+2*35.45)=0.048
Then you take your Q or quantity of heat (3570) and divide it by the number of moles in order to get the molar heat, so 3570/0.048 = 74375, or 74kJ
Since the temperature rose the reaction is exothermic, therefore -74kJ
The book says -75kJ, but that's because they use calcium as 40.1 and chlorine as 35.5 for their atomic weights. (74835.28=-75kJ)
Hope that eases your frustrations!
HJL

 

[Gussy Booo]

Ahh! Fantastic ! I'll learn from this! hahah! Eee! I forgot the valency....

and you considered the 5.3! Very nice work !

Gotta love chemistry !

Gotta hate forgetting !

 

[Kittikhun]

Damn you guys are way ahead of my class.

Anyway, I got a simple question which I don't get and which I'm not going to piss off my teacher again with.


Explain the use of the following solutions

a). barium chloride solution to test for the sulfate ion

b). silver nitrate solution to test for the chloride ion in solution

Thanks.