dec synoptic prep Flashcards
Citric acid, C3H5O(COOH)3, occurs naturally in many fruits and can also be synthesised in the laboratory for use as a food flavouring. A student analysed a sample of citric acid to determine its percentage purity.
The student dissolved 784 mg of impure citric acid in water to prepare 250 cm3 of solution in a volumetric flask.
The student titrated 25.0 cm3 samples of this solution with 0.0500 mol dm–3 sodium hydroxide solution using phenolphthalein as the indicator.
C3H5O(COOH)3(aq) + 3NaOH(aq) → C3H5O(COO)3Na3(aq) + 3H2O(l)
(a) The student rinsed the burette before filling it with the sodium hydroxide solution.
State why the student should use sodium hydroxide solution rather than water for the final rinse of the burette.
use of water would dilute the NaOH OR
use of water would change the concentration of NaOH OR
to ensure the concentration of the NaOH is not changed OR
Ignore reference to weakening the solution, watering down the solution, contaminate
Allow
it would gives a titre value that is larger
it would decrease the pH of the NaOH
A student is provided with a 5.60 g sample of ethanoic acid (CH3COOH) contaminated with sodium ethanoate (CH3COONa).
The student dissolves the sample in deionised water and makes the volume up to 200 cm3
The student removes 25.0 cm3 samples of the solution and titrates them with 0.350 mol dm–3 with volume of 20.275 cm sodium hydroxide solution.
(b) The student rinses the burette with deionised water before filling with sodium hydroxide solution.
State and explain the effect, if any, that this rinsing will have on the value of the titre
1
M2 Amount of NaOH = 0.35 × (20.275 ÷ 1000) = 0.00709625 mol
Amount of ethanoic acid in 25 cm3 = 0.00709625 mol
M2 = M1 × 10–3 × 0.35
1
M3 Amount of ethanoic acid in 200 cm3 = 0.05677 mol
M3 = M2 × 8
1
M4 Mass of ethanoic acid in sample = 60.0 × 0.05677 = 3.4062 g
M4 = M3 × 60.0
1
M5 Mass of sodium ethanoate = 5.6 – 3.4062 = 2.1938 g
M5 = 5.6 – M4
1
M6 percentage CH3COONa = (2.1938 ÷ 5.6) × 100 = 39.1 %
M6 = (M5 ÷ 5.6 ) × 100
(39.1 – 39.2)
Accept alternative methods
M5 = (M4 ÷ 5.6) × 100) followed by M6 = 100 – M5 1
This question is about a titration.
A student dissolves an unknown mass of sodium hydroxide in water to make 200 cm3 of an aqueous solution.
A 25.0 cm3 sample of this sodium hydroxide solution is placed in a conical flask and is titrated with 0.150 mol dm–3 19.575cm sulfuric acid.
The equation for this reaction is shown.
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
(b) The student uses a funnel to fill the burette with sulfuric acid before starting the titration. After filling, the student forgets to remove the funnel from the top of the burette.
A) calc mass of NaOH
Suggest why this might affect the titre volume recorded.
b)State one advantage of using a conical flask rather than a beaker for the titration.
M2 Moles of H2SO4 = concentration x volume
= 0.150 x (19.575 / 1000)
(= 2.936 x 10–3 mol)
M2 = M1 x 10–3 x 0.150
M3 Moles of NaOH in 25 cm3 = 2.936 x 10–3 x 2 = (5.87 x 10–3 mol)
M3 = M2 x 2
M4 Moles of NaOH in original 200 cm3 sample = 5.87 x 10–3 x 8
(= 0.04698 mol)
M4 = M3 x 8
M5 Mass of NaOH = Mr x moles = 40.0 x 0.04698
= 1.88 g (1.9 g)
M5 = 1.879g
b)Less chance of splashing/losing any solution using a conical flask (when swirling)
Citric acid, C3H5O(COOH)3, occurs naturally in many fruits and can also be synthesised in the laboratory for use as a food flavouring. A student analysed a sample of citric acid to determine its percentage purity.
The student dissolved 784 mg of impure citric acid in water to prepare 250 cm3 of solution in a volumetric flask.
The student titrated 25.0 cm3 samples of this solution with 0.0500 mol dm–3 and 23.95cm3 of sodium hydroxide solution using phenolphthalein as the indicator.
C3H5O(COOH)3(aq) + 3NaOH(aq) → C3H5O(COO)3Na3(aq) + 3H2O(l)
Find the mass, in mg, of citric acid dissolved in 250 cm3 of the solution.
The relative molecular mass (Mr) of citric acid is 192.0 (3)
Calculate the percentage purity of this sample of citric acid. (1)
Time of flight (TOF) mass spectrometry is an important analytical technique.
(d) In the TOF mass spectrometer, a germanium ion reaches the detector in
4.654 × 10–6 s
The kinetic energy of this ion is 2.438 × 10–15 J
The length of the flight tube is 96.00 cm
The kinetic energy of an ion is given by the equation KE=O.5 x M x V2
where
m = mass / kg
v = speed / m s–1
The Avogadro constant L = 6.022 × 1023 mol–1
Use this information to calculate the mass, in g, of one mole of these germanium ions.
Use your answer to state the mass number of this germanium ion.
M1 v = length/t = 0.96 / 4.654 × 10–6
v = 206274 m s–1
m = 2KE/v2
M1 = working (or answer)
M2 mass of one ion = 1.146 × 10–25 kg
M2 = answer conseq on M1
M3 mass of 1 mole ions = 1.146 × 10–25 × 6.022 × 1023 = (0.06901 kg)
M3 = M2 × 6.022 × 1023
M4 = 69(.01) g
M4 = M3 × 1000
M3/M4 could be in either order
M5 mass number = 69
M5 must have whole number for mass no
desribe electron spray
write equation for C3H602N undergoing electron spray
b)In the TOF mass spectrometer, a germanium ion is elector ionized deduce the equation
Sample is) dissolved (in a volatile solvent)
Allow named solvent (eg water/methanol)
1
(Injected through) needle/nozzle/capillary at high voltage/positively charged
Each molecule/particle gains a proton/H+
Ignore pressure
C3H6O2N + H+ –> C3H5O2NH+/C3H602N+
b)Ge(g) → Ge+(g) + e–
A sample of hydrated nickel sulfate (NiSO4.xH2O) with a mass of 2.287 g was heated to remove all water of crystallisation. The solid remaining had a mass of 1.344 g.
(a) Calculate the value of the integer x.
Show your working.
0.943 g water (M1)
If Mr of NiSO4 wrong, can allow M1 and M3 from method 1 i.e. max 2
NiSO4 =1.334/154.8 =8.68 × 10−3=1
H2O=0.943/18 =0.052) = 6
(M2) (M3)
x = 6 (M4)
Allow Mr = 155
Mr (NiSO4) = 58.7 + 32.1 + 64.0 = 154.8
n(NiS04XH2O)= 2.287/8.68 × 10−3= 263.4
AO 18X=108.6/18 = 6
A sample of hydrated nickel sulfate (NiSO4.xH2O) with a mass of 2.287 g was heated to remove all water of crystallisation. The solid remaining had a mass of 1.344 g.
Suggest how a student doing this experiment could check that all the water had been removed.
re-heat
Heat to constant mass = 2 marks
1
check that mass is unchanged
In an experiment a flask is used for a combustion reaction.
Method
- Remove all the air from the flask.
- Add 0.0010 mol of 2,2,4-trimethylpentane (C8H18) to the flask.
- Add 0.0200 mol of oxygen to the flask.
- Spark the mixture to ensure complete combustion.
- Cool the mixture to the original temperature.
The equation is
C8H18(g) + 12 O2(g) → 8 CO2(g) + 9 H2O(l)
Calculate the amount, in moles, of gas in the flask after the reaction. (2)
M1 amount of CO2 formed in flask = 0.008 mol
Allow ECF from M1 to M2
1
M2 amount of gas in flask
= 0.0075 (O2) + 0.0080 (M1) = 0.0155 mol
The diagram represents two glass flasks, P and Q, connected via a tap.
Flask Q (volume = 1.00 × 103 cm3) is filled with ammonia (NH3) at 102 kPa and 300 K. The tap is closed and there is a vacuum in flask P.
Calculate the mass of ammonia in flask Q.
Give your answer to the appropriate number of significant figures.
(Gas constant R = 8.31 J K−1 mol−1
B) When the tap is opened, ammonia passes into flask P. The temperature decreases by 5 °C. The final pressure in both flasks is 75.0 kPa.
Calculate the volume, in cm3, of flask P.
n = PV/RT
If PV=nRT rearranged incorrectly then M3 only
Mass = M2 × 17 = 0.696 (g) (3 sig figs only)
Allow 0.695 or 0.697
B)If pV = nRT
Incorrect unit conversion loses M1 only; can get M2/M3 if possible volume obtained
= 1.34 × 10−3 m3
Inserts correct numbers (inc pressure in Pa)
1
Volume of Q in m3 = 1.00 × 10−3
Volume of bulb P = 1.34 × 10−3 – 1.00 × 10−3
Volume of bulb P = 3.42 × 10−4 m3
No subtraction M1 only
= 342 cm3 (Allow 310 − 342 cm3)
Alternative method also worth full credit
(note if mol in M2 of 05.1 rounded to 0.04 this could lead to a final answer of 3.1 × 10−4 m3 so allow range 310 − 342 cm3
A student attempted to reduce a sample of 2-methylbutanal but added insufficient NaBH4
The student confirmed that the reduction was incomplete by using a chemical test.
Give the reagent and observation for the chemical test.
Tollens’ (reagent) OR ammoniacal silver nitrate OR description of making Tollens’
1
Silver mirror/ppt OR black solid / precipitate / deposit
1
NOT dichromate
For Tollens’ reagent:
for M1 ignore either AgNO3 or [Ag(NH3)2+] or “the silver mirror test” on their own, or “Tolling’s reagent”, but mark on
OR Fehling’s/ Benedict’s (solutions)
red solid / precipitate (allow orange or brown)
For Fehling’s/Benedict’s solution:
for M1 Ignore Cu2+(aq) or CuSO4 or “Fellings” on their own, but mark on
