J l (NaCIy 0.2 N.u J 250 Na = 23. Cl = 35.5)s Q1 :- prepare a standard solution of a Sodium Chloride (NaCl), 0.2 N in 250 ml of water. (atomic weight : Na=23 , Cl=35.5 )?
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- One mole of a component A and twomoles of a component B are mixed at270C to form an ideal binary solution. calculate the Vmix , Gmix Hmix and Smix. Assume that R=8.314 JK1mol-1 ?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 vapor pressure of the solution to 3 decimal places in atm.
- 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)
- An experiment was conducted to determine the effect of glucose on the freezing point of water. Experimental data showed thatthe freezing point depression of water in solution was –2.6°C when 1.0 g of glucose was dissolved in 10g of solvent. Calculatethe expected freezing point for such solution and compare the expected freezing point to the value found experimentally. Give aplausible explanation for any discrepancies. (M.W. of glucose= 180.16g/mol; i=1). Show all your work.A 15.0% by weight solution was prepared using 90.0g of KCl and the resulting density of the solution is 1.101g/mL. (MW KCl 74g/n). volume of solution(mL)? milliosmole of solute? %w/v? weight of solvent(g? N?A solution is prepared by dissolving 40.00 g of MgCl2 (f.w. = 95.211 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 2 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 freezing point of the solution.
- A solution is prepared by dissolving 40.00 g of MgCl2 (f.w. = 95.211 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 2 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 osmotic pressure in atm.Table 2. Gibbs Free Energies of formation (kJ), ∆G°f, for Ions in 1M Solution and Ionic Solids Cations Cl--131.228 I--51.57 NO3--108.74 SO4-2-744.53 Ca2+-553.58 -748.1 -528.9 -743.07 -1797.28 W2 ∆G°f of water = -237.129 kJ/mol Calculated values of ∆G°rxn and the ∆Grxn of each box, Predicted results (ppt or no ppt).Observations (Rxn or No Rxn). S or support and R for Refute Cations Cl- I- NO3- SO4-2 Ca+2 helpCalculate the osmolarity in mOsm/L of 500 mL of a solution containing 5% w/v dextrose, 0.78% w/v sodium chloride, and 27mEq of potassium acetate. dextrose: MW 180, # particles on dissociation 1 sodium chloride: MW 58.5, # particles on dissociation 2, valence 1 potassium acetate: MW 98, # particles on dissociation 2, valence 1