Last Assessment in F6 Chemistry Practical - EXP 20

Experiment 20

Aim: To investigate the effect of solvent volume and number of extractions on the % of extracted product

Results:
Assume the volume of KA2/ cm3 = a(i) = x ; a (ii) = y ; a (iii) = z

Questions:

(c) 2NaOH (aq) + C2H­4(COOH)2 (l) → C2H4(COONa)2 (aq) + 2H2O (l)


Ma (10)/ 0.05(x) = ½
Let concentration of C2H4(COOH)2 = a1 x 118 (Mr) = b1 g dm-3
Therefore the mass of succinic acid in 25 cm3 = b1 x 0.025 dm3 = m1 g


(d) Ma (10)/ 0.05 (y) = ½
Let concentration of C2H4(COOH)2 = a2 x 118 (Mr) = b2 g dm-3
Therefore the mass of succinic acid in 25 cm3 = b2 x 0.025 dm3 = m2 g


(e) Total mass = m1 + m2 = m3 g

(f)
Ma (10)/ 0.05(z) = ½
Let concentration of C2H4(COOH)2 = a3 x 118 (Mr) = b3 g dm-3
Therefore the mass of succinic acid in 50 cm3 = b3 x 0.05 dm3 = m3 g

(f) Smaller volume of extracting solvent will extract smaller amount of succinic acid
(g) Less volume of solvent used will be more efficient than using single larger volume, proven by total mass obtained by comparing 50 cm3 in a (i) and a (iii).

Latest update: Experiment 17

Exp 17
Topic: Technique - synthesis

Purpose: to determine the percetage of aluminium in a sample X by means of the preparation of a complex compoun dof aluminium wiht 8-hydroxyquinoline.

Procedure:

Let's assume that the result of step (a) obtained as below:

mass of container + KA3/ g = x
mass of empty container/ g = y
mass of KA3/ g = x-y = z

Results:

Let's the result of step (b) as below:

mass of crucible + precipitate/ g = a
mass of empty crucible/g = b
mass of precipitate/g = a - b = c

Question:
(c) (i) to maximised the collision between particles and thus ensures a complete reation
(ii) to remove impurities and excess ions. Cold water ensures that no precipitate will dissolve and be washed away because water with low temperature will not dissove the precipitate.

(d) the complex precipitate obtained is dried and weight a few times in the oven until the weight of the complex precipitate obtained is constant.

(e) Mass of 1 mole of complex compound with formula Al(C9H6NO)3 = 459.0 g

(f) Percetage of aluminium in the complex compound = 27.0 / 459.o x 100% = 5.88%

(g) weight of aluminium produced = 5.88/100 x c = m1 g

therefore, the percetage of Al in sample X = m1/z x 100% = p %


All the best!

CHEMISTRY FOR STPM SCHOOL-BASED ASSESSMENT :

PRACTICAL CHEMISTRY

EXPERIMENT 1: Equilibrium and solubility

The following result and answers are used as guidelines for calculation and solution for the question of (c) to (g)

V0 = volume of Pb(NO3)2
V = volume of KI

Note: Kxp is not affected by the change of concentration used in each experiment.

Try to get almost constant reading from KA1 to KA4.

(e) The values of the ionic product of PbI2 are almost constant for all the experiments.
(f) Average reading = (c1 + c2 + c3 + c4) / 4
(g) Increasing the concentration of lead (II) ions do not affect the value of Ksp of PbI2.

CHEMISTRY FOR STPM

SCHOOL-BASED ASSESSMENT : PRACTICAL CHEMISTRY

EXPERIMENT 10: Ionic Equilibrium

The following result and answers are used as guidelines for calculation and solution for the question of (c) to (g)

Results:
(b) Record and complete your readings in the table below:









(d) Dilution decreases the concentration and hence increases the pH
(e) Dilution decreases the concentration and hence increases the degree of dissociation
(f) as shown in the table above.
(g) Dilution has no effect on Ka.

Precaution step:
the bulb of pH meter must be immersed in the buffer solution before next measurement is being carried out.
measurement of pH should can done from the most diluted acid to the most concentrated acid in order to attain more accurate reading of pH.

Practical assessment of STPM Chemistry

Check out the experiment! Will be updated from time to time....

Experiment 6 (Thermochemistry)

Purpose: to determine the hear of neutralisation of a strong acid with a strong base

Questions: c) heat released of each of the experiment, q = m x c x change of T = (10 + 30) x 4.2 x change of T

d) no. of moles of sollutions can be found by n = MV/1000

e) Ionic equations: H+ + OH- --H2O

f) Heat of neutralisation: H = c)/ d)

g) acid W is a diprotic acid h) - Mixture of solution must be sitrred through out the exp. - solutions must be mixed quickly to attain the maximum change of temperature.

Experiment 12 Question Answer provided served as reference,

e) From the graph of EO against lg [Zn2+], the value of Eo increases as the value of lg [Zn2+] decreases. Therefore the e.m.f. of the cell increases when the concentration of Zn2+ ions decreases.

f) (i) with reference of Nernst Equation, 0.059 [Zn2+] Ecell = EØcell + n lg [Cu2+] The value of lg [Zn2+]/[Cu2+] increases when the concentration of Cu2+ ions decreases. Hence the e.m.f. value decreases with the decrease in Cu2+ ions.

(ii) The Eo value for Al is larger than the Eocell value for zinc. Hence with the reference to the Nernst Equation, the Eocell value will be larger/