Solubility of KNOs 70 Answer the following 3 questions by using this solubility graph of KNOs. 60 50 40 30 20 10 O 10 20 30 40 s0 60 70 80 90 Temperature (C) a) What happens to a solution of KNO; that is saturated at 50°C when it is cooled quickly to 10°C? EXPLAIN. b) How would you describe a solution of KNO, at 80°C if there are about 45 g of KNO, dissolved in 100 g of water? EXPLAIN. (8 0OL6) Aamanjos
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- The following evidence was obtained from an experiment to determine the solubility of calcium chloride at room temperature. A sample of saturated calcium chloride solution was evaporated to dryness, and the mass of solid residue was measured.EvidenceVolume of solution (mL) = 15.0Mass of empty beaker (g) = 90.54Mass of beaker and residue (g) = 101.36The solubility of calcium chloride is g/100 mL1.1The Ksp of Ca3 (PO4 ) 2 is 1.3 × 10−26 . Estimate the solubility of this salt in units of g. L −1 . You must show any reaction equation(s) that you may think are necessary. 1.2 If a sample of solid Ca3(PO4)2 is stirred into exactly one litre of a 0.550M solution of Na3PO4, how will the solubility of the salt compare with the answer that you have obtained in question 1.1? Explain you answer in a short sentence.A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the freezing point of the solution(in C to 2 decimal places)
- A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the osmotic pressure in atm to three decimal placesA solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the boiling point of the solution(in C to 2 decimal places)A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the following: Boiling point of solution (in °C to two decimal places) Freezing point of solution (in °C to two decimal places) Vapor pressure of the solution (in atm to three decimal places) Osmotic pressure (in atm to three decimal places)
- 1. The Ksp of Ca3(PO4)2 is 1.3 × 10−26. Estimate the solubility of this salt in units of g. L−12. If a sample of solid Ca3(PO4)2 is stirred into exactly one litre of a 0.550M solution of Na3PO4, how will the solubility of the salt compare with the answer that you have obtained in question 2.1? Explain you answer in a short sentence.Calculate the solubility at 25°C of CuBr in pure water and in a 0.0100M CoBr2 solution. You'll find Ksp data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: gL solubility in 0.0100 M CoBr2solution: gLA 1.00-m solution of acetic acid, CH3COOH, in benzene has a freezing point of 2.96°C. Use the data in the Table to calculate the value of i and suggest an explanation for the unusual result. (Hint: If i is less than 1.0, each formula unit that dissolves yields less than one solute particle, an outcome suggesting aggregation of solute particles.) Answer is: i = 0.50; formation of dimers of composition (CH3COOH)2, need steps shown to understand though
- Develop the solubility diagrams for CuCO3 in a closed system and an open system, respectively. Please provide rationale or workings for your answer.Calculate the solubility of Ba(IO3)2 in pure water, Ksp=1.5x10^-9. Then calculate the solubility of the same Ba(IO3)2 in a 0.0025M Al(NO3)3 solution. Assume there is no other interatcion between any of the ionic componets of the 2 molecules. (do an activity coeffiecient problem using the neatest u value to determine f from the table)The compound lead(II) sulfate has a solubility of 4.824E-3 g/100mL at 25 degC.What is the molar solubility of this solution at saturation? 0.0002 mol/L What is the Ksp at 25 degC for lead(II) sulfate? What is the molar solubility of lead(II) sulfate in 0.28 M lead(II) nitrate ?