All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet
(a) What difference in physical properties enable the separation of mixtures by:
(i) simple distillation,
(ii) paper chromatography;
(iii) fractional distillation.
(b) Give a reason for each of the following practices during titration in the laboratory.
(i) White tile is placed under the conical flask.
(ii) Burette readings are always recorded to two decimal places.
(iii) Calculate the volume of 2.5 moldm\(^{-3}\) stock HČI required to prepare 500 cm\(^3\) of 0.20 moldm HCI.
(a) Physical property differences that allow separation:
- (i) Simple distillation: a difference in boiling points, one component being volatile (a liquid) and the other non-volatile (e.g. a dissolved solid), so the liquid boils off and is condensed.
- (ii) Paper chromatography: a difference in solubility and adsorption, i.e. the components dissolve to different extents in the solvent and are adsorbed to different extents on the paper, so they travel at different rates.
- (iii) Fractional distillation: a difference in the boiling points of two (or more) miscible liquids whose boiling points are fairly close together; the fractionating column allows them to be separated.
(b) Reasons for titration practices:
- (i) A white tile is placed under the conical flask so that the colour change at the end point can be seen clearly against a white background.
- (ii) Burette readings are recorded to two decimal places for accuracy/precision, because the burette is graduated to 0.1 cm\(^3\) and readings are estimated to the nearest 0.05 cm\(^3\).
(iii) Volume of 2.5 mol dm\(^{-3}\) stock HCl needed to prepare 500 cm\(^3\) of 0.20 mol dm\(^{-3}\) HCl. Using \(C_1V_1 = C_2V_2\):
\[2.5 \times V_1 = 0.20 \times 500\]
\[V_1 = \frac{0.20 \times 500}{2.5} = \frac{100}{2.5} = 40\,cm^3\]
So 40 cm\(^3\) of the stock acid is required, then made up to 500 cm\(^3\) with distilled water.
(a) Physical property differences that allow separation:
- (i) Simple distillation: a difference in boiling points, one component being volatile (a liquid) and the other non-volatile (e.g. a dissolved solid), so the liquid boils off and is condensed.
- (ii) Paper chromatography: a difference in solubility and adsorption, i.e. the components dissolve to different extents in the solvent and are adsorbed to different extents on the paper, so they travel at different rates.
- (iii) Fractional distillation: a difference in the boiling points of two (or more) miscible liquids whose boiling points are fairly close together; the fractionating column allows them to be separated.
(b) Reasons for titration practices:
- (i) A white tile is placed under the conical flask so that the colour change at the end point can be seen clearly against a white background.
- (ii) Burette readings are recorded to two decimal places for accuracy/precision, because the burette is graduated to 0.1 cm\(^3\) and readings are estimated to the nearest 0.05 cm\(^3\).
(iii) Volume of 2.5 mol dm\(^{-3}\) stock HCl needed to prepare 500 cm\(^3\) of 0.20 mol dm\(^{-3}\) HCl. Using \(C_1V_1 = C_2V_2\):
\[2.5 \times V_1 = 0.20 \times 500\]
\[V_1 = \frac{0.20 \times 500}{2.5} = \frac{100}{2.5} = 40\,cm^3\]
So 40 cm\(^3\) of the stock acid is required, then made up to 500 cm\(^3\) with distilled water.
