Using the value of 1.853 degrees C/m for Kf, de rmine the difterence In (JATE (C a Cl2 solution) minus ATf (N a Cl solution)| (absolute value) if each solution contains 0.19 moles of the solute and 489.00 grams of water. The solutes are C a Cl2 and N a CI. Use ATf = i Kf m for each solute. Your Answer: Answer
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- In class, we derived the expression:ln XA =(ΔvH/R) ((1/T)-(1/T°))where A represents the solvent (and B would represent the solute, such that XA + XB = 1).Using the approximations given below, derive the expression:∆Tb = XB (RT°2/ΔvH)where ∆Tb is the change in boiling point between the solution and the pure liquid, T − T°.Approximations:ln(1 − X) ≈ −XTT° ≈ T°2Note: this derived expression is often simplified in General Chemistry as ∆Tb = Kbm where Kb is the boiling point elevation constant, and m is the molality.calculate the sulablity Iat 25 Celsius of Ni(OH)2 pure water and then .01 80 M NaOH solution. Ksp of Ni(OH)2 is 5.48x10^16 additional info in the picture5-7: A 10.34 molal aqueous nitric acid solution was prepared. (Density of nitric acid = 1.51 g/mL, density of water = 1.0 g/mL, and the MW of nitric acid = 63 g/mol)5. How many grams of HNO3 were used to prepare the solution? 6. What is the Molarity of the solution? 7. What is the concentration (%v/v) of the solution? 8-10: An aqueous solution of HNO3 was prepared having a molal concentration of 6.29. (Atomic weights: H=1 amu, N=14 amu, O=16 amu, density of H2O = 1g/mL, density of HNO3 = 1.51 g/ml)8. How many grams of HNO3 was used to prepare the said concentration? 9. What is the molarity of the aqueous HNO3 solution? 10. The % v/v of the prepared HNO3 solution is __________.
- What will be the new boilng point of 498ml benzene (Kb = 2.53 oC•Kg/mol; Bp = 80.1oC) when mixed with 18g of naphthalene (MW = 128.17 g/mol) to make a non- eletrolyte solution? Density of benzene = 0.876 g/ml)An excess amount of Ba(NO3)2 was added to 2 liters of water at 25 oC until the solution is saturated. Because Ba(NO3)2 is only slightly soluble in water, the excess Ba(NO3)2 was filtered out to get a solid-free mixture. The proponents of the experiment thought about using the concept of colligative properties, specifically boiling point elevation, to determine Ksp. It was observed in their experiments that the solution boils at 100.15 deg C. Kb of water = 0.51 K-kg/mol. Barium nitrate dissociates using: Solve the ff: a) Amount of Ba(NO3)2 dissolved in grams. MW of Ba(NO3)2 = 261.3 g/molb) Solubility product, Kspc) Vapor pressure of the solution in kPaThe distribution coefficient, KD (C ether/C water), between ether and water for aspirin at room temperature is 3.5. What weight of aspirin would be extracted by a single extraction with 150 mL of water from a solution of 10 gms of aspirin in 100 mL of ether? Calculate the weight of aspirin which would be removed by three extractions with 50 mL portions of water.
- What would be the boiling point of 500 mL of chloroform if 10.50 grams of caffeine (molar mass = 194.19) was added? The normal boiling point of chloroform is 61.3oC, the density of chloroform is 1.49g/mL, and the Kb of chloroform is 3.63 oC/m. Only enter the numerical value for the answer. DO NOT include a unit.1. How many microliters of a 25 \mu M MgCl2.6H2O solution would be needed to provide 152.48 ng of MgCl2.6H2O for a chemical reaction? [ Mwt MgCl2.6H2O = 203.3 g/moleAn experimental attempt was to convert 2000 cm^3 of a solution consists of 0.96 ethanol into another solution consisting of 0.56 ethanol. If the density of water at the experiment's temperature and pressure is 997 kg/ m^3, How much H2O should be added to the 2000 cm^3 solution? What is the solution's volume after conversion? Given data: In the 0.96 ethanol solution :V ethanol = 0.816 cm^3/g and Vwater = 1.273 cm/g In the 0.58 ethanol solution: V ethanol = 0.953 cm^3/g and :V water = 1.243 cm/g
- Use the following steps to determinr the percent error of the freezing point depressions you determined experimentally from the expected freezing point depression for each solute. Show sample calculations for sodium chloride solution. a. Use the actual molar mass of each solute to determine the expected motality of 0.234g of each solute in 4.00x10-3kg of water. Sample Calculation: b. Determine the expected freezing point depression for a solution with this motality. Sample Calculation: c. Determine the percent error of the freezing point depression you determine experimentally from the expected freezing point depression for each solute. Sample Calculation:A solution is prepared by dissolving iodine, I2, in carbon tetrachloride, CCl4, at 25 °C. What is the mole fractions of the solute for a solute of molality 0.100 mol kg−1? R = 8.3145 J/Kmol. Mole fraction of solute = 0.015 Mole fraction of solvent = 0.985 (c) change of chemical potential (solvent) = __________ J/mol. 3 sig. fig.Given the following data for: Mass of test tube and stearic acid = 14.17 g Mass of test tube = 11.40 g Freezing point of stearic acid = 69.59 oC Mass of weighing paper + naphthalene = 1.230 g Mass of weighing paper = 0.920 g Freezing point solution = 64.00oC Kf = 4.5 oC/m Determine the following: mass of stearic acid in g (2 decimal places): mass of naphthalene in g (2 decimal places): freezing point depression (2 decimal places):