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Frage 1 Bericht
Burette readings (initial and final reading) must be given to two decimal places. Volume of pipette used must also be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is 0.0950 mol dm\(^{-3}\) HCI. B is a solution 13.50g dm\(^{-3}\) of X\(_2\)CO\(_3\).10H\(_2\)O.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions öf B using methyl orange as an indicator. Tabulate your readings and calculate the average volume of A used.
(b) From your results and the information provided above, calculate the;
(i) concentration of B in mol dm\(^{-3}\);
(ii) molar mass of X\(_2\)CO\(_3\).10H\(_2\)O in g mol\(^{-1}\);
(iii) percentage by mass X in X\(_2\)C)\(_3\).10H\(_2\)O. [H = 1, C = 12, O = 16]. The equation for the reaction involved in the titration is 2HCl\(_{(aq)}\) + X\(_2\)CO\(_3\).10H\(_2\)O\(_{(aq)}\) \(\to\) 2XCl\(_{(aq)}\) + 11H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(a) Titration
A (0.0950 mol dm-3 HCl) is placed in the burette and 25.0 cm3 portions of B are pipetted into the conical flask with methyl orange as indicator. A specimen set of concordant readings is shown.
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|---|---|---|
| Final reading | 25.10 | 24.90 | 24.85 |
| Initial reading | 0.00 | 0.00 | 0.00 |
| Volume of A used | 25.10 | 24.90 | 24.85 |
Average volume of A (concordant titres only) \(=\dfrac{24.90+24.85}{2}=24.88\ \text{cm}^3\)
(b)(i) Concentration of B in mol dm-3
\[ n(\text{HCl})=\frac{0.0950\times 24.88}{1000}=2.364\times10^{-3}\ \text{mol} \]From the equation, HCl : X2CO3.10H2O = 2 : 1, so
\[ n(B)=\frac{2.364\times10^{-3}}{2}=1.182\times10^{-3}\ \text{mol in }25.0\ \text{cm}^3 \] \[ [B]=\frac{1.182\times10^{-3}\times1000}{25.0}=0.0473\ \text{mol dm}^{-3} \](ii) Molar mass of X2CO3.10H2O
Since concentration in g dm-3 = molar mass \(\times\) concentration in mol dm-3:
\[ M=\frac{13.50}{0.0473}=285.4\approx 286\ \text{g mol}^{-1} \](iii) Percentage by mass of X in X2CO3.10H2O
Mass of the (CO3 + 10H2O) part \(= 12 + (3\times16) + 10(18) = 60 + 180 = 240\).
So mass of 2X \(= 286 - 240 = 46\), giving X = 23 (sodium).
\[ \%X=\frac{46}{286}\times100=16.1\% \]The titre figures above are specimen readings; use your own concordant burette values. The result is consistent with X = Na, i.e. washing soda, Na2CO3.10H2O.
Antwortdetails
(a) Titration
A (0.0950 mol dm-3 HCl) is placed in the burette and 25.0 cm3 portions of B are pipetted into the conical flask with methyl orange as indicator. A specimen set of concordant readings is shown.
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|---|---|---|
| Final reading | 25.10 | 24.90 | 24.85 |
| Initial reading | 0.00 | 0.00 | 0.00 |
| Volume of A used | 25.10 | 24.90 | 24.85 |
Average volume of A (concordant titres only) \(=\dfrac{24.90+24.85}{2}=24.88\ \text{cm}^3\)
(b)(i) Concentration of B in mol dm-3
\[ n(\text{HCl})=\frac{0.0950\times 24.88}{1000}=2.364\times10^{-3}\ \text{mol} \]From the equation, HCl : X2CO3.10H2O = 2 : 1, so
\[ n(B)=\frac{2.364\times10^{-3}}{2}=1.182\times10^{-3}\ \text{mol in }25.0\ \text{cm}^3 \] \[ [B]=\frac{1.182\times10^{-3}\times1000}{25.0}=0.0473\ \text{mol dm}^{-3} \](ii) Molar mass of X2CO3.10H2O
Since concentration in g dm-3 = molar mass \(\times\) concentration in mol dm-3:
\[ M=\frac{13.50}{0.0473}=285.4\approx 286\ \text{g mol}^{-1} \](iii) Percentage by mass of X in X2CO3.10H2O
Mass of the (CO3 + 10H2O) part \(= 12 + (3\times16) + 10(18) = 60 + 180 = 240\).
So mass of 2X \(= 286 - 240 = 46\), giving X = 23 (sodium).
\[ \%X=\frac{46}{286}\times100=16.1\% \]The titre figures above are specimen readings; use your own concordant burette values. The result is consistent with X = Na, i.e. washing soda, Na2CO3.10H2O.
Frage 2 Bericht
Credit will be given for strict adherence to the instructions, for observations precisely recorded and for accurate inferences. All tests, observations and inferences must be clearly entered in your answer book in ink, at the time they are made.
C is a mixture of two salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion you draw from the result of each test.
(a) Put all of C in a test tube and add about 10 cm\(^3\) of distilled water. Shake thoroughly and filter. Keep both the filtrate and the residue. Divide the filtrate into three portions.
(i) To the first portion, add NaOH\(_{(ag)}\) in drops and then in excess.
(ii) To the second portion, add NH\(_3\), solution in drops and then in excess.
(iii) To the third portion, add BaCl\(_{2(aq)}\) followed by dilute HCI.
(b) Divide the residue into two portions.
(i) Heat the first portion strongly in a test tube.
(ii) Add dilute HCI to the second portion.
| Test | Observation | Inference |
|---|---|---|
| C was shaken with distilled water and filtered. | A blue filtrate and a green residue were obtained. | C contains a soluble salt and an insoluble salt. |
| (a)(i) To the first portion of the filtrate, NaOH(aq) was added dropwise and then in excess. | A blue gelatinous precipitate formed. It was insoluble in excess NaOH(aq). | Cu2+ ions are present. |
| (a)(ii) To the second portion of the filtrate, NH3(aq) was added dropwise and then in excess. | A pale-blue gelatinous precipitate formed. It dissolved in excess ammonia solution to give a deep-blue solution. | Cu2+ ions are confirmed. |
| (a)(iii) To the third portion of the filtrate, BaCl2(aq) was added, followed by dilute HCl. | A white precipitate formed and remained insoluble in dilute HCl. | SO42− ions are present. |
| (b)(i) The first portion of the residue was heated strongly. | The green solid changed to a black powder. A colourless gas was evolved; it turned limewater milky. | The residue contains a trioxocarbonate(IV). The gas is carbon(IV) oxide, CO2. |
| (b)(ii) Dilute HCl was added to the second portion of the residue. | Effervescence occurred. A colourless, odourless gas was evolved and turned limewater milky. | The gas is carbon(IV) oxide, CO2; hence CO32− is present. |
The soluble salt is copper(II) tetraoxosulphate(VI), CuSO4, while the green insoluble salt is copper(II) trioxocarbonate(IV), CuCO3. Therefore, C is a mixture of CuSO4 and CuCO3.
The relevant reactions are:
\[\mathrm{CuCO_3(s) \xrightarrow{heat} CuO(s) + CO_2(g)}\]
\[\mathrm{CuCO_3(s) + 2HCl(aq) \rightarrow CuCl_2(aq) + H_2O(l) + CO_2(g)}\]
Antwortdetails
| Test | Observation | Inference |
|---|---|---|
| C was shaken with distilled water and filtered. | A blue filtrate and a green residue were obtained. | C contains a soluble salt and an insoluble salt. |
| (a)(i) To the first portion of the filtrate, NaOH(aq) was added dropwise and then in excess. | A blue gelatinous precipitate formed. It was insoluble in excess NaOH(aq). | Cu2+ ions are present. |
| (a)(ii) To the second portion of the filtrate, NH3(aq) was added dropwise and then in excess. | A pale-blue gelatinous precipitate formed. It dissolved in excess ammonia solution to give a deep-blue solution. | Cu2+ ions are confirmed. |
| (a)(iii) To the third portion of the filtrate, BaCl2(aq) was added, followed by dilute HCl. | A white precipitate formed and remained insoluble in dilute HCl. | SO42− ions are present. |
| (b)(i) The first portion of the residue was heated strongly. | The green solid changed to a black powder. A colourless gas was evolved; it turned limewater milky. | The residue contains a trioxocarbonate(IV). The gas is carbon(IV) oxide, CO2. |
| (b)(ii) Dilute HCl was added to the second portion of the residue. | Effervescence occurred. A colourless, odourless gas was evolved and turned limewater milky. | The gas is carbon(IV) oxide, CO2; hence CO32− is present. |
The soluble salt is copper(II) tetraoxosulphate(VI), CuSO4, while the green insoluble salt is copper(II) trioxocarbonate(IV), CuCO3. Therefore, C is a mixture of CuSO4 and CuCO3.
The relevant reactions are:
\[\mathrm{CuCO_3(s) \xrightarrow{heat} CuO(s) + CO_2(g)}\]
\[\mathrm{CuCO_3(s) + 2HCl(aq) \rightarrow CuCl_2(aq) + H_2O(l) + CO_2(g)}\]
Frage 3 Bericht
(a) List three pieces of apparatus required for the evaporation of sodium chloride solution to dryness.
(b)(i) List two normal salts which when dissolved in water turn red litmus blue.
(ii) State the phenomenon that is responsible for the action on the litmus in (b)(i).
(c) State what would be observed on adding BaCl\(_2\) solution to a portion of a saturated Na\(_2\)CO\(_3\), followed by dilute HCI in excess.
(i) A gas Q decolourized acidified KMnO\(_4\) solution. Suggest what Q could be.
(a) Three pieces of apparatus for evaporating sodium chloride solution to dryness
Evaporating dish (basin), tripod stand and Bunsen burner. (Wire gauze and a glass stirring rod are also acceptable.)
(b)(i) Two normal salts that turn red litmus blue
Sodium trioxocarbonate(IV), \(Na_2CO_3\), and potassium trioxocarbonate(IV), \(K_2CO_3\). (Sodium ethanoate, \(CH_3COONa\), is also acceptable.) Their aqueous solutions are alkaline.
(b)(ii) Phenomenon responsible
Salt hydrolysis (specifically anionic hydrolysis): being salts of a strong base and a weak acid, they hydrolyze in water to give an alkaline solution.
(c) Adding BaCl2 to saturated Na2CO3, then excess dilute HCl
On adding barium chloride solution, a white precipitate of barium trioxocarbonate(IV) forms:
\[BaCl_{2(aq)} + Na_2CO_{3(aq)} \to BaCO_{3(s)} + 2NaCl_{(aq)}\]On then adding dilute hydrochloric acid in excess, the white precipitate dissolves with effervescence, giving off a colourless gas (carbon(IV) oxide) that turns limewater milky:
\[BaCO_{3(s)} + 2HCl_{(aq)} \to BaCl_{2(aq)} + H_2O_{(l)} + CO_{2(g)}\](i) Gas Q that decolourizes acidified KMnO4
Q is a reducing gas; it could be sulphur(IV) oxide (sulphur dioxide, \(SO_2\)). (Hydrogen sulphide, \(H_2S\), or an unsaturated gas such as ethene would also decolourize acidified \(KMnO_4\).)
Antwortdetails
(a) Three pieces of apparatus for evaporating sodium chloride solution to dryness
Evaporating dish (basin), tripod stand and Bunsen burner. (Wire gauze and a glass stirring rod are also acceptable.)
(b)(i) Two normal salts that turn red litmus blue
Sodium trioxocarbonate(IV), \(Na_2CO_3\), and potassium trioxocarbonate(IV), \(K_2CO_3\). (Sodium ethanoate, \(CH_3COONa\), is also acceptable.) Their aqueous solutions are alkaline.
(b)(ii) Phenomenon responsible
Salt hydrolysis (specifically anionic hydrolysis): being salts of a strong base and a weak acid, they hydrolyze in water to give an alkaline solution.
(c) Adding BaCl2 to saturated Na2CO3, then excess dilute HCl
On adding barium chloride solution, a white precipitate of barium trioxocarbonate(IV) forms:
\[BaCl_{2(aq)} + Na_2CO_{3(aq)} \to BaCO_{3(s)} + 2NaCl_{(aq)}\]On then adding dilute hydrochloric acid in excess, the white precipitate dissolves with effervescence, giving off a colourless gas (carbon(IV) oxide) that turns limewater milky:
\[BaCO_{3(s)} + 2HCl_{(aq)} \to BaCl_{2(aq)} + H_2O_{(l)} + CO_{2(g)}\](i) Gas Q that decolourizes acidified KMnO4
Q is a reducing gas; it could be sulphur(IV) oxide (sulphur dioxide, \(SO_2\)). (Hydrogen sulphide, \(H_2S\), or an unsaturated gas such as ethene would also decolourize acidified \(KMnO_4\).)
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