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The boiling point of an aqueous solution is 101.45 Celsius. What is the freezing point?
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- Some nonelectrolyte solute (molar mass = 142 g/mol) was dissolved in 150. mL of a solvent (density = 0.879 g/cm3). The elevated boiling point of the solution was 355.4 K. What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 kJ/mol, the entropy of vaporization is 95.95 J/K mol, and the boiling-point elevation constant is 2.5 K kg/mol.Boiling Point Elevation/Freezing Point Depression T = m K where, for freezing point depression: T = T(pure solvent) - T(solution) and for boiling point elevation: T = T(solution) - T(pure solvent) m = (# moles solute / Kg solvent) Kb = boiling point elevation constant. Kf = freezing point depression constant. Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow. Solvent Formula Kb(°C / m) Kf(°C / m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02What is ΔSsys for a freezing phase transition at 29.7 °C for a compound that freezes at 29.7 °C and the ΔHsys = -19.45 kJ mol-1 for this process? Express your answer in J mol-1 K-1 to at least two significant figures.
- The table below shows temperature/composition data collected for a mixture of methylbenzene (M) and octane (O) at 1 atm. Recall that x stands for the mole fraction in the liquid and y stands for the mole fraction in the vapor in equilibrium. The boiling points for methylbenzene (M) and octane (O) are 110.60C and 125.60C, respectively. Construct the phase diagram with Temperature vs. xM. What is the composition of the vapor in equilibrium with the liquid of composition (a) xM = 0.250 and (b) xO = 0.250. T (0C) 110.9 112.0 114.0 115.8 117.3 119.0 121.1 123.0 xM 0.908 0.795 0.615 0.527 0.408 0.300 0.203 0.097 yM 0.923 0.836 0.698 0.624 0.527 0.410 0.297 0.164At 39.9 ◦C, a mixture of ethanol (x1 = 0.9060, p1 = 130.4 Torr) and isooctane ( p2 = 43.9 Torr) forms a vapor phase with molar fraction y1 = 0.6667 at a total pressure of 185.9 Torr. (a) Calculate the activity and activity coefficient of each component. (b) Calculate the total vapor pressure that the mixture would have if it were ideal.The freezing-point of depression of a 0.010 b acetic acid solution is 0.00193 K. Calculate thedegree of dissociation for acetic acid at this concentration. Cryoscopic constant for the aceticacid is 1.86 K mol kg mol-1.
- Extra information: freezing point depression constant for the solvent (20.0 degrees Celcius kg/mol for cyclohexane)Estimate the freezing point of 250 cm3 water sweetened by the addition of 2.7 g sucrose. Treat the solution as ideal. 0.068 0C 0.059 0C -0.068 0C -0.059 0CSuppose that a solution made from an unknown solute dissolved in solvent X. Its molality is 0.0125 and it boils 0.047oC higher than solvent X's boiling point. With the use of valid approximations, calculate the osmotic pressure of the said solution at 60oC? The following are useful thermodynamic data for solvent X to solve the problem: ΔSvap=87.92 J*K-1mol-1 Tb=61.15oC density at 60oC=1.394 g/cm3
- A solution of common salt in water is prepared by adding 50kg of salt to 200kg of water tomake a liquid density of 1250kg/m3. Calculate the concentration of salt in this solution as a: i) Weight(w/w)fraction (Ans: weight fraction 20%)Phenol and water form non-ideal liquid mixtures. When 7.32 g of phenol and 7.95 g of water are mixed together at 60 °C they form two immiscible liquid phases with mole fractions of phenol of 0.042 and 0.161. (i) Calculate the overall mole fraction of phenol in the mixture. (ii) Use the lever rule to determine the relative amounts of the two phases.Calculate the boiling point (Tbp) of an aqueous solution that is 2.222molal sucrose(aq)under standard conditions. Report the value with 4 sig figs.