what is the osmotic pressure of a 0.250 M K2SO4 solution at 45 degrees celsuis?
817492
Ok so the equation for osmotic pressure is:\n\npi = iMRT\npi = Osmotic Pressure\ni = i-factor\nM = Molarity\nR = Gas Constant (0.08211 L*atm / mol*K)\nT = Temperature (in kelvin)\n\nNow you know MRT because of what they give you but you need to find the value of the i-factor. The i-factor is the van 't hoff factor and is equivalent to the number of molecules associated or dissociated from the original solute. Since K2SO4 dissociates into 2 K+ ions and 1 SO4 2- ion, then the i-factor is 3.\n\nTherefore, plug in the numbers and solve.\n\npi = 3(0.250 mol/L)(0.08211 L*atm / mol*K)(318K)\npi = 19.6 atm
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
307097
Ksp = [Ag+]^2[CrO4 2-]\nsince [Ag+] = 2*[CrO4 2-]\nKsp = 2*[Ag+]^3\nsolve for [Ag+] then plug it in this equation\n[Ag+] = mass /0.010L H2O
849074
K2S + COCl2 --> 2KCl + COS
354566
The osmolarity property.
761477
Actually you have an electrolyte, so the drop in the freezing point  is\n\nΔT=i*Kf*m\n\nwhere i is the van't Hoff coefficient and (for simplicity) for strong electrolytes it is equal to the number of ions that are produced by its dissociation, according to the molecular formula.\nNaCl dissociates completely to two ions (Na+ and Cl-) thus i=2\n\nfor water Kf=1.86\n\nm is the molaLity and not molaRity. \nYour solution is 2 M = 2mole solute/L of solution \nmolality is mole solute/kg of solvent. You need additional info in order to convert molarity to molality\n\nIf you had 2 molaL and not molaR solution then m=2, or if we assume due to lack of more info that in this case molaRity=molaLity...\n\nΔT=i*Kf*m =2*1.86*2 = 7.44\n For pure water Tfwater=0 C, so for the solution \nTf= Twater-ΔT= 0-7.44= -7.44 C
19338
We will assume that 2.0 M NaCL(aq) is a dilute solution.  Therefore, 1 L of water weighs 1 kg.  For concentrated solutions, such as acids, you need to know the density of the solution.\n\nTo know the boiling point, use the formula\nDT = Kb * m\n\nWhere DT = change in temperature\nKb = boiling point constant (0.52 for water) \nm = molality.\n\nFirst, calculate the molality of the the solution\n\nMolality = moles of solute per kiograms of solven\n\n\n2.0 moles of NaCl / 1 kg solvent = 2.0\n\nDT=Kb * m\n= .52 * 2.0\n= 1.04 degrees\n= 1.0 degrees (using 2 significant digits)\n\nSo, water boils at 100 degrees C, so the boiling point of 2.0M NaCl is 101 degrees C.\n\nHope it helps.\n\n-Mark
569971
This problem only has enough information if you assume it is an ideal gas.  In this case, PV=nRT is the equation of state.\n\nn and R are constants in this case, so PV/T must also be constant.  These values must also all be in absolute scales - the celcius must be converted to kelvin or any other scale based on absolute zero.  This means:\n\nP(i)V(i)/T(i)=P(f)V(f)/T(f)\nV(f)=T(f)P(i)V(i)/[P(f)T(i)]\n    =(250+273)K(740mmHg)(410mL)/[(680mmHg)(27+273)K]\n    =[523*740*410/(680*300)]mL\n    =777.83mL\n\nBy the significant digits you've presented, this can only be evaluated to any accuracy as 780mL.  (two significant digits)
711424
The Equation should be:\nNaCl + H2SO4 = Na2SO4 + HCl\nSo the coefficients are (2,1,1,2)
764442
2KClO3(s) --> 2KCl(s) + 3O2(g)\n\nmole of KClO3 = 0.263 / 122.6 = 2.145 mmole\nmole of O2 = 2.145 * 3 / 2 = 3.217 mmole\nvolume of O2 over water:\nP * V = n * R * T\n0.948 atm * V = 3.217 * 10^-3 * 0.082 * 294\nV = 0.081 litre\n\npeace\nvixklen
170129
CuSO4 (s)  --H2O-->  Cu+2 (aq) + SO4-2 (aq)
918146
CuSO4.5H2O => CuSO4.H2O + 4H2O at 98*C (or 110*C)\nCuSO4.H2O => CuSO4 + H2O at 120*C (or 150*C)\nCuSO4 => CuO + SO2 + O2  at 650*C
797125
The heat lost by the warmer water will equal the heat gained by the cooler water.  The change in the heat will equal \nq = (s)(m)(T1 - T2)   where q is the heat change (units of Joules (J)), s is the specific heat of water (4.184 J/gK), m is the mass of the water in grams, and T1 and T2 are the initial and final temperatures of the water samples in Kelvin.  Assume the density of water = 1.00g/mL.\n\n(4.184J/gK)(30g)(Tf - 280K)  =  (4.184J/gK)(50g)(330K - Tf)\nTf is the final temperature of the combined solutions.\n\nSolving for Tf:\nTf = [50(330) + 30(280)] / 80  = 311K
479787
In 100g of solution that is 5% by mass NaCl, there are 95 g H2O and 5g Nacl.\nYou want 250g of solution...set up a simple ratio:\n95g H2O/100g Soln = ?g H2O/250g Soln\n? = 237.5g H2O
823368
Of what?\n\nπ= i*C*R*T\n\nT= absolute temperature (in Kelvin)\nR= the gas constant in whatever units you need to express osmotic pressure (e.g. if you want π in atm then R=0.082 L*atm/(mole*K))\nC= the concentration of your solute in mole/L\n\nand...\n\ni is the van't Hoff coefficient.\nFor non-electrolytes i=1\nFor strong electrolytes i= the number of ions that are produced by the dissociation according to the molecular formula e.g for NaCl you have 2 ions (1 Na+ and 1 Cl-) so i=2, for CaCl2 3 ions (1 Ca+2 and 2 Cl-) so i=3.\nFor weak electrolytes, if n is the number of ions coming from the 100% dissociation according to the molecular formula and a the degree of dissociation then i=(1-a)+na. E.g. if we assume for CH3COOH a=80% i=(1-0.8)+2*0.8= 0.2+1.6=1.8
366921
Q = mc delta T\nHeat = mass times specific heat times change in temperature\nspecific heat for water is 4.186 kJ/kg C\n-41840 J (negative because absorbed) = .200 kg x 4.186 x 10^3 J/ kg C x (Tf - 20)\nTf = final Temp.\nsolving the equation gives you Tf = -30.0 degrees Celsius
791970
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K