While diluting a solution of salt in water, a student by mistake added acetone (boiling point `56^(@)`). What technique can be employed to get back the acetone? Justify your choice.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point `56^(@)`). What technique can be employed to get back the acetone? Justify your choice.
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To recover acetone from a solution where it has been mistakenly mixed with saltwater, the following steps can be taken using the technique of simple distillation:
### Step-by-Step Solution:
1. **Understanding the Mixture**: The mixture consists of water, salt, and acetone. Acetone is a volatile liquid with a boiling point of 56°C, while water has a boiling point of 100°C. Salt does not affect the boiling point significantly as it dissolves in water.
2. **Choosing the Right Technique**: Since acetone is soluble in water, a separating funnel cannot be used to separate them. Instead, simple distillation is the appropriate technique because it allows for the separation of liquids based on their boiling points.
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Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 . Given : Freezing point depression constant of water (K_(f)^("water")) = 1.86 K "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethonal")) = 2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water")) = 0.52 K kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethonal")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethonal = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethonal = 40 mm Hg Molecular weight of water = 18 g "mol"^(-1) Molecular weight of ethonal = 45 g"mol"^(-1) In answering the following questions, consider the solution to be ideal ideal solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such lthat the molecules fraction of water in t he solution becomes 0.9 . The boiling point of this solution is :
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 . Given : Freezing point depression constant of water (K_(f)^("water")) = 1.86 K "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethonal")) = 2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water")) = 0.52 K kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethonal")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethonal = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethonal = 40 mm Hg Molecular weight of water = 18 g "mol"^(-1) Molecular weight of ethonal = 45 g"mol"^(-1) In answering the following questions, consider the solution to be ideal ideal solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such lthat the molecules fraction of water in t he solution becomes 0.9 . The boiling point of this solution is :
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The boiling point of an aqueous solution of a non-volatile solute is 100.15^(@)C . What is the freezing point of an aqueous solution obtained by dilute the above solution with an equal volume of water. The values of K_(b) and K_(f) for water are 0.512 and 1.86^(@)C mol^(-1) :
The boiling point of an aqueous solution of a non-volatile solute is 100.15^(@)C . What is the freezing point of an aqueous solution obtained by dilute the above solution with an equal volume of water. The values of K_(b) and K_(f) for water are 0.512 and 1.86^(@)C mol^(-1) :
Properties such as boiling point, freezing point and vapour, pressure of a pure solvent change Propeties such as boiling point, freezing point and vapour, pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 Given Freezing point depression constant of water (K_(f)^("water"))=1.86 K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol"))=2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water"))=0.52 kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol"))=1.2 kg "mol"^(-1) Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water =373 K tandard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mmHg Vapour presure of pure ethanol =40g Hg Molecular weight of water =18 g"mol"^(-1) Molecules weight of ethanol =46 g "mol"^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 The freezing point of the solution M is
Strong acid versus strong base: The principle of conductometric titrations is based on the fact that during the titration, one of the ions is replaced by the other and invariable these two ions differ in the ionic conductivity with the result that thhe conductivity of the solution varies during the course of the titration. take, for example, the titration between a strong acid, say HCl, and a string base, say NaOH before NaOH is added, the conductance of HCl solution has a high value due to the presence of highly mobile hydrogen ions. As NaOH is added, H^(+) ions are replaced by relatively slower moving Na^(+) ions. consequently the conductance of the solution decreases and this continues right upto the equivalence point where the solution contains only NaCl. Beyond the equivalence point, if more of NaOH is added, then the solution contains a excess of the fast moving OH^(-) ions with the result that its conductance is increased ad it condinues to increase as more and more of NaOH is added. If we plot the conductance value versus the amount of NaOH added, we get a curve of the type shown in Fig. The descending portion AB represents the conductances before the equivalence point (solution contains a mixture of acid HCl and the salt NaCl) and the ascending portion CD represents the conductances after the equivalence point (solution contains the salt NaCl and the excess of NaOH). The point E which represent the minium conductance is due to the solution containing only NaCl with no free acid or alkali and thus represents the equivalence point. this point can, however, be obtained by the extrapolation of the lines AB and DC, and therefore, one is not very particular in locating this point expermentally as it is in the case of ordinary acid-base titrations involving the acid-base indicators. Weak acid versus strong base: Let us take specific example of acetic acid being titrated against NaOH . Before the addition of alkali, the solution shows poor conductance due to feeble ionization of acetic acid. Initially the addition of alkali causes not only the replacement of H^(+) by Na^(+) but also suppresses the dissociation of acetic acid due to the common ion Ac^(-) and thus the conductance of the solution decreases in the beginning. but very soon the conductance start increasing as addition of NaOH neutralizes the undissociated HAc to Na^(+)Ac^(-) thus causing the replacement of non-conducting HAc with Strong-conducting electrolyte Na^(+)Ac^(-) . the increase in conductance continuous right up to the equivalence point. Beyond this point conductance increases more rapidly with the addition of NaOH due to the highly conducting OH^(-) ions, the graph near the equivalence point is curved due to the hydrolysis of the salt NaAc . The actual equivalence point can, as usual, be obtained by the extrapolation method. In all these graphs it has been assumed that the volume change due addition of solution from burrette is negnigible, hence volume change of the solution in beaker the conductance of which is measured is almost constant throughout the measurement. Q. The nature of curve obtained for the titration between weak acid versus strong base as described in the above passage will be:
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