Chapter Tests Flashcards
0.0019 L is equivalent to:
1.9 mL
100 mg is equivalent to:
10–4 kg
78 mg L–1 is equivalent to:
78 ppm
4.0 ppm is equivalent to:
4.0 μg mL–1
4.5%(w/w) KF is equivalent to
0.0077 mol L–1
When would you use a quantitative analysis ?
when determining the sodium content in mineral water
Which best describes a electrical conductivity analytical method?
The measurement of combined level of dissolved salts
Atomic absorption spectroscopy is suitable for the detection of many metals. A solution of an unknown metal is compared with solutions of known concentration. The most likely unit of concentration used in such analyses would be
ppm (Parts per million)
Aluminium can be produced by the reaction between alumina, Al2O3, and carbon. The equation for the reaction is
2Al2O3(l) + 3C(s) → 3CO2(g) + 4Al(l)
What would result in the production of 0.8 mol of aluminium?
0.4 mol of Al2O3 and 0.6 mol of C
Calculate the mass of water that is produced when 2.80 g of propane, C3H8, is completely burned in air.
C3H8 + 5O2 → 3CO2 + 4H2O
n(C3H8) = m/M = 2.80/44.0 = 0.0636 mol Therefore,
n(H2O) = 4 × 0.0636 = 0.254 mol and m(H2O)
= n × M =
0.254 × 18.0 = 4.57 g.
What mass of water will be produced if 25.0 g of hydrogen gas, H2, and 100.0 g of oxygen gas, O2, are mixed and ignited?
2H2(g) + O2(g) → 2H2O(l)
n(H2) = m/M = 25.0/2.0 = 13 mol and n(O2) = m/M = 100.0/32.0 = 3.13 mol
H2 is in excess (3.13 mol of O2 reacts with 2 × 3.13 = 6.26 mol of H2).
Therefore, O2 is the limiting reactant.
n(H2O) = 2 × n(O2) =
m/M = 2 × 3.13 = 6.26 mol
m(H2O) = n × M =
6.26 × 18.0 = 113 g
A standard solution is a solution that has
an accurately known formula.
Primary standard must
Be readily obtainable in pure form, have a known formula and not absorb water from the atmosphere
To prepare a standard solution, the set of apparatus normally used is
A wash bottle, an electronic balance and a volumetric flask.
A titre is a volume measured using a
Burette
An aliquot is a volume measured using a
Pipette
A 25.0 mL sample of vinegar was diluted to 200.0 mL in a volumetric flask.
The ethanoic acid, CH3COOH, in a 25.00 mL aliquot of this diluted solution required the addition of 23.75 mL of 0.147 M sodium hydroxide for a complete reaction. The reaction that occurred was
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
(Determine the molarity (M) (concentration in mol L–1) of ethanoic acid in the diluted vinegar
solution.)
- n(NaOH) = c × V = 0.147 × 0.023 75 = 0.003 49 mol -
- n(CH3COOH) in 20.00 mL aliquot = 0.003 49 mol
-c(CH3COOH) in 20.00 mL aliquot =
n/V = 0.003 49/0.025 00 = 0.140 M
A 25.0 mL sample of vinegar was diluted to 200.0 mL in a volumetric flask.
The ethanoic acid, CH3COOH, in a 25.00 mL aliquot of this diluted solution required the addition of 23.75 mL of 0.147 M sodium hydroxide for a complete reaction. The reaction that occurred was
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
(Determine the molarity (concentration in mol L–1) of ethanoic acid in the undiluted vinegar solution.)
concentration(CH3COOH) in 250 mL volumetric flask = 0.140 M
-Since the original vinegar solution was diluted by a factor of 200.0/25.0 = 8 8 c(CH3COOH) in the undiluted vinegar = 8 × 0.140 = 1.12 M.
A 25.0 mL sample of vinegar was diluted to 200.0 mL in a volumetric flask.
The ethanoic acid, CH3COOH, in a 25.00 mL aliquot of this diluted solution required the addition of 23.75 mL of 0.147 M sodium hydroxide for a complete reaction. The reaction that occurred was
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
(Express the concentration of ethanoic acid in the undiluted vinegar as %(m/v).)
-1.12 M means that there is 1.12 mol of CH3COOH in 1 L of solution.
-Since m = n × M 60 = Mass of CH3COOH
= 1.12 × 60.0 = 67.2 g of CH3COOH in 1 L of solution and 6.72 g of CH3COOH in 100 mL of solution.
- The concentration = 6.72%(m/v).
(Na2CO3.xH2O). A 0.300 g of this sample reacts completely with 20.0 mL of 0.250 M hydrochloric acid according to the equation
Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
(Determine the percentage by mass of sodium carbonate, Na2CO3, in the washing powder.)
n(HCl) = c × V
= 0.250 × 0.0200 = 0.005 00 mol
n(Na2CO3) = 0.5 × n(HCl) =
0.002 50 mol m(Na2CO3) =
n × M = 0.0250 × 106.0 = 0.265 g %Na2CO3 =
(0.265/0.300) × 100 = 88.3%
The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL
(Write a balanced equation with states for the reaction)
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL
(Describe the function of the methyl orange indicator)
to detect the end point
The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL
(Determine the volume of the titre, in mL, of H2SO4).
V(H2SO4) =
21.48 – 2.32 = 19.16 mL
The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL
(Calculate the amount in moles of sulfuric acid used in the titration)
n(H2SO4)=c×V=
- 09560×0.01916=
- 001832mol
The results of an analysis using methyl orange as an indicator follows.
Concentration of sulfuric acid, H2SO4 = 0.095 60 M Aliquot of sodium hydroxide, NaOH = 25.00 mL
Initial volume of sulfuric acid, H2SO4 = 2.32 mL
(Calculate the concentration in molL–1 of the sodium hydroxide solution)
n(NaOH) = 2 × n(H2SO4) = 2 × 0.001 832 = 0.003 664 mol c(NaOH) = n/V = 0.003 664/0.025 00 = 0.1466 M
