How do you find solubility of of solvents?
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The Merck Index and the Handbook of Chemistry and Physics are good resources for this.
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It won't matter how much you started with or how much higher the solution was heated.  When the solution cools to 20C it will hold 0.22g/100mL of solvent (water).\n\nSo if you start with 1.75g and 0.22g remain in solution, the amount crystallized at 20C will be the difference, 1.53g.
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Go here to find out about  "MOLAR ENTHALPY"\n\nhttp://www.carolina.com/chemistry/activities/Determining_Molar_Enthalpy_of_Solution.pdf
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use beers law to figure out the concentration and then take that to mole ratio
1125157
molality = moles of solute / kg of solvent\n\n.  Osmolality:\n*  One mole of non ionic solute dissolved in a kilogram of water will yield Avogadro's number (6.02 x 1023) of molecules.\n*  One mole of ionic solute (e.g., NaC I) dissolved in a kilogram of water will yield almost twice as many particles. The colligative properties are changed almost twice as much.\n*  Osmolality =  n molality \n*   = osmotic coefficient, or percent deviation from "ideal" behavior (complete dissociation, no water binding, etc.). \n*  n =  number of particles into which the molecule can be broken (1 for sucrose, 2 for NaC 1, 3 for K2SO4, etc.).\n*  One osmol of any solute plus one kilogram of water will yield Avogadro's number of particles.\n*  Avogadro's number of particles, regardless of their size or shape, will change the colligative properties the amount described in 3a. (when dissolved in a kilogram of water).\n\n\n Converting other units to milliosmolality (mOsm/kg):\n\nmEq/ldivide by wc\nmultiply by \ndivide by v\n\nppmdivide by wc\nmultiply by \ndivide by v\ndivide by atomic weight (aw)\n\nmg%divide by wc\nmultiply by \ndivide by mw\nmultiply by n\nmultiply by 10\n\nmillimolaritydivide by wc\nmultiply by \nmultiply by n\n\nmillimolalitymultiply by \nmultiply by n\n\nmillidegrees Cdivide by 1.86\n\nExplanation of symbols:\nwc = percent water content of solution, or, (wet weight   dried weight) / wet weight\n = osmotic coefficient\nv = valence\naw = atomic weight\nmw = molecular weight\nn = number of particles into which the molecule can be broken.
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1. Make sure that HCl is standardized.\n2. Write the balance equation:\n\nCa(OH)2 + 2HCl ---> CaCl2 + 2H2O\n\n- this means that 1 mole of Ca(OH)2 reacts for every 2 moles of HCl\n\n3. Get the number of moles of HCl reacted by multiplying the volume used to react Ca(OH)2 by the actual concentration of HCl (value from standardization). Divide the result by 2 - and this is now your moles Ca(OH)2.\n\n4. Get the molar solubility of Ca(OH)2 by dividing its number of moles by its volume. Unit is in gram per liter.
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(1) Take a container and measure the length of the container. Let us say the length of the container is L. \n\n(2) Take a beam of light (preferably a monochromatic source like laser beam) and shine it through the empty container. \n\n(3) Measure the intensity of the light at the end where the beam has gone through the empty container. Let this be Io. \n\n(4) Now fill the container with the liquid whose molar extinction coefficient is unknown. Now measure the light intensity as the beam travels the length L of the container. The measured intensity is I.\n\n(5) Beer-Lamberts law is I=Io exp(-a L) where all the quantities have been experimentally determined except a which is the extinction coefficient. \nRearrange the above formula and we get, a=(-1/L)*log(I/Io) where log is to the base e.
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The molarity of a solution is the number of moles of a substance per liter of solution.\n\nMolarity = moles/liter\n\nSo, if you know what your solute is, you need to calculate the number of moles you have by:'\n\nmoles = mass of solute/molecular weight of solute\n\nnumber of moles is then divided by the number of liters you have, and voila!  Molarity.\n\nExample:  40 grams of sodium chloride in 500 mL of water:\n\n40 gm NaCl/58.443 grams per mole = 0.6844 moles\n\n0.6844 moles/500 mL * 1000 mL/liter = 1.369 moles/liter  or the solution is 1.369 M sodium chloride in water.
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You can calculate the freezing point depression of the solution is you know three things, firstly, the freezing point of the pure solvent, and secondly, the concentration of the solute, and thirdly, the freezing point depression constant of the solvent.\nIn this problem, you know both (One is given, two of them are assumed knowledge).\n\nThe freezing point of pure water is 0 degrees Celcius.\nThe concentration of the solute is 52 percent.  You need to convert this to a molaltiy concentration to use it (or fiind a different freezing point depression constant).\nLook up for your self the k_f value.\n\nthe freezing point depression = k_f * m,\nwhere k_f is the freezing point depression and m is the concentraion of the solute in terms of molality.\n\nmolality = moles of solute / kg is solvent
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It has to do with the "activiation energy" involved in the process, increasing the stirring rate in a certain manner, increases the rate of collisions which increases the rate of reactions.  Solubility has to do with the free energy which are state functions with respect to temperature; the reactant and product states. Rates of reactions pertain more towards transitions states.
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Unfortunately, this question can not be solved because it lacks pertinent information that is required to solve the problem.  But, I will lead you through the steps as if all the information were present.\n\nThe first step would be to determine how many moles of substance is present, the temperature(in Kelvins), and the volume(in L), using the R value of .0821.  This plugs into the equation PV=nRT.  Solve for P, which gives the total pressure of the system. \n\nThe next step is to determine the mole fraction for each of the substances.  This is accomplished by dividing the number of moles of the particular substance by the total moles used earlier.  This value is the mole fraction.\n\nFrom here, to find the partial pressure for the particular substance, multiply the mole fraction by the total pressure of the system.  This will give you the partial pressure for that substance.\n\nThen, using Dalton's Law of Partial Pressures, subtract the partial pressure value from the total pressure to determine the partial pressure of the other substance.  Now, the problem is solved.  Thus, the substance with the greatest mole fraction will have the greatest partial pressure.
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To work out the volume of solvent needed to make a 10mM (0,01M) solution use the formula:\n\nKnown mass of solid\n---------------------------------     = volume of solvent needed (in litres)\n0.01  x  Relative Mass of solid
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The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
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1) Decide what solvent you want to use: water or organic?\n2) Decide how much solution you want: 1L?\n3) Check the solubility of phenol in the solvent you choose. And calculate how much phenol you'll need by weight to make a saturated solution\n4) Find out the relationship between solubility and temperature.\n5) I assume you want to make a saturated phenol solution at room temperature. If the solubility increases as temperature goes higher, add a bit more phenol than needed (from the calculation above) to the solvent. \n6) Heat the flask or beaker that contains the solvent and phenol until all phenol dissolves.\n7) Let the container cool down to room temperature.\n8) If no phenol crystalizes the solution is over-saturated. You may rub the side of the flask or beaker to help phenol crystalize.\n9) You have your saturated phenol solution!!
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The CRC Handbook of Chemistry and Physics has all kinds of good information like that (although it is pricey).\n\nOr, search the web for the specific liquid of interest.
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If you know the volume and density of the liquid you can calculate the number of moles n.\n\nOnce you have n and if you assume that the gas will behave like and ideal gas and you fix the temperature and pressure, the volume will be given by:\n\nV = nRT/P.
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Most solutes (sugar in this case) have limited solubility (ability to dissolve) in most solvents (water in this case). This solubility varies with temperature - the hotter the solution, the higher the solubility. As mentioned in an earlier answer, this means more energy must be added to dissolve more solute. On the other hand, if energy is taken out of a solution, the solute will usually  come out of the solution, depending on the concentration.\n\nThis physical law that governs this is the 2nd law of thermodyamics.