i need Help with questions 14 and 15 from the 2009 chemistry hsc exam. (1 Viewer)

[Riot09]

Help with questions 14 and 15 from the 2009 chemistry hsc exam.



(for the titration question i use c1v1=c2v2 and I'm not used to the other method.)

the link for the paper is:

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/pdf_doc/2009-hsc-exam-chemistry.pdf

 

[findsome1]

Question 14: Avoid using C1V1= C2V2, instead write the equation for the reaction out, then use given data to find out moles and concentrations.

Question 15:
By the graph, theres is 8mg L of O2 concentration at 25 degrees. Convert this to grams first so it is 8x10^-3 g/L.
Then find the number of moles using n= m/M where m is 8x10^-3 and M is 32 (for O2)
THis should then give you 2.5x10^-3 moles.

Then use the formula n= V/24.79 (because its at 25 degress and standard pressure)

so when using the formula 2.510^-3 = V/24.79
the, according to the calculations
V= 61.97 ml, which is approx 62.0 mL, so the answer is A

 

[s2 SEductive]

Help with questions 14 and 15 from the 2009 chemistry hsc exam.



(for the titration question i use c1v1=c2v2 and I'm not used to the other method.)

the link for the paper is:

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/pdf_doc/2009-hsc-exam-chemistry.pdf

c1v1=c2v2 is used for dilution...

q14. 3H+ + OH- --> h20 Since citric acid is triprotic , its 3 times [H]+

What we know:
Concentration of NaOH: 0.550 mol L-1
Volume of titraton (NaOH): 29.50mL = 0.02950 L
Volume of solution to be titrated( citric acid) = 25.0mL = 0.025L
Molar mass of Citric Acid = 192.12g mol-1

What we need to find:
Concentration of Citric acid

How?
1. We must find the number of moles of NaOH, by C=n/v

n(NaOH) = 0.550 x 0.02950
= 0.016225 mol
3H+ + OH- = h20

n(H+..Citric acid) = 0.016225/3 {since 3H+ : 1OH-)

n(citric acid) = 0.005408333

C(citric acid) = 0.005408333/0.025 {Volume(citric acid) = 0.025L)}
= 0.216 mol L-1
Concentration in g L-1
You must multiply by molar mass of citric acid

0.216 x 192.12 = 41.56

= 41.6 g L-1 ( correct to 1.d.p)


q15

From the graph we know that at 25 degrees , only 8mg of oxygen will dissolve,

In 10 L , 8x10 = 80mg will dissolve

n(O2) = 80/32.0 n=m/M
= 2.5x10^-3
V(O2) = 2.5x10^-3 x 24.79 [Volume of 1 mole ideal gas at 100kPa at 25 degrees is 24.79 L
= 0.062L = 62.0mL