ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Tea.bag on February 24, 2008, 01:09:52 am

Title: Chem Question help
Post by: Tea.bag on February 24, 2008, 01:09:52 am: I am having trouble answering the following questions. Can anyone help me? :o

1. To what volume of water must 10ml of 8.0M HCL be added in order to prepare a 0.50M HCL solution?

2. The chlorine in a 0.63g sample of a chlorinated pesticide, DDT (C₁₄H₉Cl₅), is precipitated as silver chloride. What mass of silver chloride is formed?
Title: Re: Chem Question help
Post by: Collin Li on February 24, 2008, 01:22:00 am: Question 1

Using $c_1V_1 = c_2V_2$

$\implies 8.0\times0.010 = 0.50V_2$

$\implies V_2 = 0.16\mbox{ L} = 160\mbox{ mL}$

Since $V_2$ tells you the new volume: volume to be added is the difference between the new volume and the old volume.

Therefore, 150 mL must be added.
Title: Re: Chem Question help
Post by: Collin Li on February 24, 2008, 01:28:23 am: Question 2

$\mbox{n(AgCl)} = \mbox{n(Cl)} = 5\times\mbox{n(C}_{14}\mbox{H}_9\mbox{Cl}_5\mbox{)} = \frac{5\times 0.63}{14\times 12.0 + 9\times 1.0 + 35.5\times 5}$

$\implies \mbox{n(AgCl)} = \frac{3.15}{354.5} = 0.0089\mbox{ mol}$

$\implies \mbox{m(AgCl)} = 0.0089(107.9+35.5) = 1.3\mbox{ g}$
Title: Re: Chem Question help
Post by: Tea.bag on February 24, 2008, 01:36:16 am: Thanks Coblin.

More questions coming soon. :P
Title: Re: Chem Question help
Post by: Tea.bag on March 13, 2008, 06:34:33 pm: Got another question...

1. The organic compounds dimethyl ether $CH_3OCH_3$ and ethanol $CH_3CH_2OH$ have the same molecular formula, $C_2H_6O$ , and the same molar mass; however they have different structural formulas.
A sample containing both ethanol and dimethyl ether is analysed by GLC using a flame ionisation detector.

a) Will the two compounds produce peaks with the same retention time? Explain you answer.
b) If the sample have the same concentration of both chemicals, will the peaks produced have the same area? Explain your answer.

2. Fluoride compounds are added to Melbourne's supplies of drinking water to give a concentration of fluoride ions of about 0.90 ppm.

a) What amount, in mol, of fluoride is present in 1.0g if Melbourne water?
b) How many fluoride ions would you swallow if you drank a 200 ml glass of Melbourne water?
Title: Re: Chem Question help
Post by: Mao on March 13, 2008, 08:07:23 pm: 1a)
no they will not
clearly seen, ethanol is more polar, and will adsorb/desorb more than dimethyl ether
1b)
yes, the area under the peak directly varies against concentration.
this is also the reason why HPLC/GC can be used quantitatively as well as qualitatively

2a)
*hint*
ppm, parts per million, literally means parts per million
90 millionth of 1.0g
2b)
you can assume 1L of water to be 1kg, or an average density of 1g/mL.
Title: Re: Chem Question help
Post by: Collin Li on March 13, 2008, 08:57:04 pm: Question 1

Quote
a) Will the two compounds produce peaks with the same retention time? Explain you answer.

It should be sufficient to say that because you have a molecule with a different structural connectivity (despite having the same chemical composition), then one of these molecules is different in chemical and physical properties to the other (and hence interacts differently with the mobile and stationary phase in GLC). You shouldn't need to state which one is more polar, although it is nice to know (use your commonsense here, the -OH group exposed on the ethanol is more polar than the C-O-C bridge that is surrounded by other atoms).

Question 2

Quote
a) What amount, in mol, of fluoride is present in 1.0g if Melbourne water?

$0.90\mbox{ ppm} = 0.90\mbox{ \mu}\mbox{g/g}$

Don't be confused by the "microgram by gram" notation. It means "one microgram" of A per "gram" of B. Scaling this up by a factor of $10^6$ reveals that this proportion is exactly the same as the "parts per million" notation: one gram of A per million grams of B. Note that "ppm" is in terms of grams - gladly assume this unless you are told otherwise.

In this case, "A" is fluoride ions and "B" is water, so using $0.90\mbox{ ppm} = 0.90\mbox{ \mu}\mbox{g/g}$ , we know that per gram of water:

$\mbox{n(F}^-\mbox{)} = \frac{0.90 \times 10^{-6}\mbox{ g}}{19.0\mbox{ gmol}^{-1}} = 4.74 \times 10^{-8}\mbox{ mol}$

Quote
b) How many fluoride ions would you swallow if you drank a 200 mL glass of Melbourne water?

Using the density of water: $1\mbox{ g/mL}$ , we can say that if there is $\mbox{n(F}^-\mbox{)} = 4.74 \times 10^{-8}\mbox{ mol}$ per gram of water, then in 200mL (200 grams) of water, there are:

$\mbox{n(F}^-\mbox{)}_{\mbox{200 mL}} = 4.74 \times 10^{-8} \times 200 = 9.47 \times 10^{-6}\mbox{ mol}$

Now, it says "how many fluoride ions," so you have to actually give the number of ions, rather than in terms of moles:

$\mbox{N(F}^-\mbox{)} = 9.47 \times 10^{-6} \times 6.02\times 10^{23} = 5.70 \times 10^{18}$
Title: Re: Chem Question help
Post by: Tea.bag on March 16, 2008, 11:00:42 pm: o_O so simple..

thanks coblin
Title: Re: Chem Question help
Post by: Tea.bag on March 31, 2008, 04:56:43 pm: question..

1) A $2L$ sample of a gaseous hydrocarbon is burnt in excess oxygen. The only products of the reaction are $8L$ of $CO_2$ (g) and $10L$ of $H_2O$ (g) all at 100 degrees celcius at 1 atm pressure. what is the formula of the hydrocarbon.

here is my working..

$n=\frac{PV}{RT}$

$n(CO_2)= \frac{101.325×8}{8.31×373}=0.26 mol$

$\implies n(C)=n(CO_2)=0.26 mol$

$n(H_2O)=\frac{101.325×10}{8.31×373}=0.33 mol$

$\implies n(H)=2×0.33=0.65 mol$

$n(C):n(H)$

$0.26:0.65$

$1:2.5$

$2:5$

$\implies C_2H_5$

but the answer is $C_4H_{10}$

can i make the assumption that $C_2H_5$ is an empirical formula and so i times that by two to get the answer? the book has a different working out aswell..plz help
Title: Re: Chem Question help
Post by: Collin Li on March 31, 2008, 06:18:22 pm: The hydrocarbon has the formula $\mbox{C}_{\mbox{x}}\mbox{H}_{\mbox{y}}$ .

You know that $\mbox{n(C}_{\mbox{x}}\mbox{H}_{\mbox{y}}\mbox{)} : \mbox{n(CO}_2\mbox{)} = 2:8 = 1:4$

Therefore, the hydrocarbon must have 4 carbons (i.e: $x=4$ ). I've sort of made some leaps in logic, but I think you can fill in the gaps. Basically because carbon dioxide only has one carbon per molecule, and we are assuming a complete combustion of the hydrocarbon, then we know that the moles of carbon in the reactants must match up with the moles of carbon in the products. If the molar ratio of the hydrocarbon to the carbon dioxide is a quarter, then that means there must be 4 times the amount of carbon per molecule of the hydrocarbon than the carbon dioxide. Does that make sense?
Title: Re: Chem Question help
Post by: Tea.bag on March 31, 2008, 07:12:21 pm: yes thnx