Mol of Solute Molality ( ) = Kg Solvent M

Home , Gas constant, Molality, Osmotic pressure

• Molality (m) is the number of moles of solute per kilogram of solvent.

mol of solute Molality (m ) = kg solvent

Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water. mol of solute m = kg solvent 1 m ol KF ()13.5g 58.1 =  0.250 kg = 0.929 m

• Mole fraction (χ) is the ratio of the number of moles of a substance over the total number of moles of substances in solution. number of moles of i χ = i total number of moles n = i nT

• ex) What is the molality of a 0.200 M aluminum nitrate solution (d = 1.012g/mL)? – Work with 1 liter of solution. mass = 1012 g

– mass Al(NO3)3 = 0.200 mol × 213.01 g/mol = 42.6 g ; – mass water = 1012 g -43 g = 969 g

0.200mol Molality==0.206 mol / kg 0.969kg

Calculate the mole fraction of 10.0g of NaCl dissolved in 100. g of water.

mol of NaCl χ=NaCl mol of NaCl + mol H2 O 1 mol NaCl ()10.0g 58.5g NaCl =  1 m ol NaCl 1 m ol H2 O ()10.0g+ () 100.g H2 O  58.5g NaCl 18.0g H2 O = 0.0299

g solute mass % or weight %=× 100 g solution gsolute ppm =×106 gsolution gsolute ppb =×109 g solution g solute ppt =×1012 gsolution

11 gsolute mgsolute ppm =≈ 106 g of solution L solution 11 g solute µ g solute ppb =≈ 109 g of solution L solution 11 g solute ng solute ppt =≈ 1012 g of solution L solution

• The solubility of a gas decreases with temperature.

• Nature of solute and solvent • Most nonelectrolytes that are appreciably soluble in water are hydrogen bonded (methanol, hydrogen peroxide, sugars). Other types of nonelectrolytes are generally more soluble in nonpolar or slightly polar solvents such as benzene or toluene.

• Effect of temperature a. Increase in T favors endothermic process: solid + water → solution ∆H usually positive, so solubility increases with T gas + water → solution ∆H usually negative, so solubility decreases with T Effect of pressure Henry’s Law Negligible, except for gases, where solubility is directly proportional to the partial pressure of the gas. Carbonated beverages.

• The solubility of an ionic solid generally increases with temperature.

Henry’s Law

Pgas = kgasCgas

–Pgas = pressure of the gas above the solution

–Cgas = concentration of the gas

–kgas = Henry’s law constant

Henry’s law holds best for gases O2 and N2, does not hold HCl

A liter of water dissolves 0.0404 g of oxygen at 25oC at a pressure of 760. torr. What would be the concentration of oxygen (in g/L) if the pressure were increased to 1880 torr at the same temperature?

PP 12= CC12 0.0404 g/L C = 2 760. torr 1880 torr

C2 = 0.0999 g/L

• Vapor pressure lowering – the vapor pressure of a solvent is lowered by the addition of a nonvolatile solute.

cf) volatile solute

• The properties of a solution differ considerably from those of the pure solvent.

The solution properties depend primarily upon concentration of solute particles rather than type.

• Vapor pressure lowering

• Boiling point elevation, freezing point lowering

• Osmosis, osmotic pressure

o Psolution= χ solventP solvent

o P solvent = vapor pressure of the pure solvent

It is independent of the nature of the solute but directly proportional to its concentration.

What will be the vapor pressure of a solution made

by dissolving 6.25g of glucose, C6H12O6 , in 50.0g of water at 25oC? How much was the vapor pressure of the pure water lowered? The vapor pressure of water at 25oC is 23.8 torr 1 mol mol glucose = 6.25 g = 0.0347 mol glucose 180. g 1 mol mol water = 50.0 g= 2.78 mol water 18.0g 2.78 χ == 0.988 water 0.0347+2.78 o Psoln== waterP water ()() 0.988 23.8 torr = 23.5 torr vapor pressureχ lowering = 23.8 torr - 23.5 torr = 0.2 torr

Boiling point elevation – the change in the boiling point is:

∆Tb = iKbm – i = sum of the coefficients of the ions (i = 1 for molecular compounds)

–Kb = boiling point elevation constant – m = molality

Freezing point depression – the change in the freezing point is:

∆Tf = iKfm – i = sum of the coefficients of the ions (i = 1 for molecular compounds)

–Kf = freezing point depression constant – m = molality

Calculate the boiling point elevation and the freezing point depression

of a solution made by dissolving 12.2g of KCl in 45.0g of water. Kb = o o 0.512 C/m and Kf = 1.86 C/m

• i = 2 for KCl → K+ + Cl−

1 mol mol KCl = 12.2g= 0.164 mol 74.6g mol KCl 0.164 mol mm== =3.64 KCl kg water 0.045 kg

oo ∆=Tbbim K =() 2() 0.512 C/ m() 3.64 m =3.73 C

() oo ∆=Tffim K = 2() 1.86 C/ m() 3.64 m =13.5 C

• Water moves through semi-permeable membrane from region of high vapor pressure (pure water) to region of low vapor pressure (solution)

• π = nRT/V = MRT

•1 M solution at 25°C has osmotic pressure of 24.5 atm

Pickle, Fresh water, Nutrient solution

Osmotic pressure П = iMRT – i = sum of the coefficients of the ions (i = 1 for molecular compounds) – M = molarity – R = gas constant (0.0821 L•atm/mol•K) – T = temperature in Kelvin

• Colligative effects are greater because of increased number of particles.

• ∆Tf = 1.86°C × m × i a. where i is approximately equal to the number of moles of ions per mole of solute: (only in very dilute solution)

• NaCl(s) → Na+(aq) + Cl-(aq) i = 2 • CaCl(s) → Ca2+(aq) + 2Cl-(aq) i = 3 a. Actually, i is usually less than predicted because of ionic atmosphere effects and ion pair.

What is the osmotic pressure of a 100. mL solution containing 9.50 g of glucose, o C6H12O6, at 20.0 C?

1 mol mol glucose = 9.50 g =0.0528 mol 180. g 0.0528 mol M = = 0.528 mol/L glucose 0.100 L Latm⋅ π = (1)() 0.528 mol/L 0.08206 () [20.0+273] K = 12.7 atm Kmol⋅