Interpretation:
The reason for the low recovery of the crystal in the experiment performed by crystallization of biphenyl needs to be explained, if the sample weigh 0.5 g with 5% impurity and after crystallization the crystal dried and weigh 0.02 g.
Concept Introduction:
Organic compounds are mainly covalent compounds in which C-C, C-H and C-X bonds are present. Here, 'X' indicates heteroatom which is present in
A polar covalent molecule will be soluble in water whereas a non-polar covalent compound will be insoluble in water based on the concept of "Like dissolves like". Some of the polar compounds are also not soluble in water as they cannot form hydrogen bonding with water molecules.
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EBK A SMALL SCALE APPROACH TO ORGANIC L
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- Consider solutions of Cr(NO3)3-9H2O in water. The solubility of Cr(NO3)3-9H2O in water at 15 °C is 208 g of Cr(NO3)3-9H2O per 100 g of water. A solution is formed at 35 °C by dissolving 303 g of Cr(NO3)3-9H₂O in 100 g water. When this solution is slowly cooled to 15 °C, no precipitate forms. What is the nature of the resulting cooled solution? Saturated Solvated Supersaturated Hydrated Unsaturatedarrow_forwardOral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample.arrow_forwardOral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample. The stated should be 104.5, how??arrow_forward
- A group was given the responsibility of performing a recrystallization experiment in their laboratory to see if the three newly found solvents could purify and separate a mixture including compounds A and B. The following are the solubilities of the two compounds in g/100 mL of hot and cold solvents: Solubility in g/100 mL solvent A B Solvent Hot Cold Hot Cold 1 4.60 1.10 5.30 1.90 2 6.40 5.80 8.40 1.10 3 9.70 3.30 6.80 1.70 What is the %recovery and %purity of A and B in the chosen solvent? Show solutions plsarrow_forwardAn ideal solvent to use for recrystallization is one that: Group of answer choices dissolves the desired compound in ice-cold water temperature but does not dissolve the desired compound at its boiling temperature. a solvent in which the desired compound dissolves at any temperature. a solvent that reacts with the desired compound. dissolves the desired compound at its boiling temperature but does not dissolve the desired compound in ice-cold temperature.arrow_forwardStudents were tasked to conduct a recrystallization experiment to test the ability of the newly discovered solvents in their laboratory to purify and separate a mixture containing compounds X and Y. The solubility of the two compounds in g/50 mL of hot and cold solvents are given by the photo.arrow_forward
- 1-propanol, 2-propanol, and 1-butanol all have boiling too close together to be separated using the simple distillation column. What equipment be used to separate a mixture of these three?arrow_forwardAn impure sample contains 0.74 g of impurities and 4.55 g of benzoic acid. The sample is dissolved in 100 mL of water and heated. The solution is then cooled. How many grams of the impurity will crystallize when the solution has cooled? Include the unit and two decimal places in your answer. The solubilities of the impurity and benzoic acid in cold and hot water are found as an "item" in the "Recrystallization" folder. Solubility Data for Report Questions 11 Compound Cold Water solubility Hot Water Solubility Benzoic acid 0.42 g per 100 mL of water 5.79 g per 100 mL of water Impurity 0.61 g per 100 mL of water 4.62 g per 100 mL of waterarrow_forwardWhy is fractional distillation more efficient than simple distillation when purifying liquid mixtures?arrow_forward
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