Solutions & t heir P hysical Properties

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# Solutions & t heir P hysical Properties - PowerPoint PPT Presentation

Solutions & t heir P hysical Properties. Solution Concentrations. Mass of solute Percent by mass = ------------------------ x 100 Mass of solution Volume of solute Percent by volume = ------------------------- x 100 Volume of solution Mass of solute

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### Solutions &their Physical Properties

Solution Concentrations

Mass of solute

Percent by mass = ------------------------ x 100

Mass of solution

Volume of solute

Percent by volume = ------------------------- x 100

Volume of solution

Mass of solute

Mass / volume percent = --------------------- x 100 (gr/100ml)

Volume of solution

Solution Concentrations

• Number of moles of solute
• Molarity = -----------------------------------------------
• Number of liters of solution
• Number of moles of solute
• Molality = --------------------------------------------------
• Number of kilograms of solvent
• Moles of component i
• Mole fraction = ---------------------------------------------------------
• total moles of all solution components
• Parts per million (ppm), Parts Per Billion(ppb)
• mg solute g solute
• 1 ppm = -------------------- 1 ppb = -----------------------
• kg solution kg solution
question.

A solution has a density of 1.235g/ml and contains 90.0% glycerol C3H8O3 and 10% H2O by mass. Determine

a- the molarity of C3H8O3

b- the molarity of H2O

c- mole fraction of C3H8O3

d- mole percent of H2O

e- the molality of H2O

Solubility of most of solids increase with temperature.
• Solubility of gases decrease with increased temperature
• Solubility of a gas increases as the gas pressure is increased.
• Types of solvent and solute is important,

like dissolves like, generally nonpolar substances are soluble in nonpolar substances, polar solvents dissolves ionic and polar substances.

Solution Process

• a- Separation of solvent molecules from one another
• Pure solvent → separated solvent molecules ∆ H > 0
• b- Separation of solute molecules from one another
• Pure solute → separated solute molecules ∆ H > 0
• c- mixing solvent & solute molecules
• separated solvent molecules
• + ↔ solution ∆ H < 0
• separated solute molecules
• ∆ H solution = ∆ H a + ∆ H b + ∆ Hc

Intermolecular Forces in Mixtures

• **Like dissolves like**
• Substances with similar intermolecular attractive forces tend to be soluble in one another.
• Polar liquids tend to dissolve in polar solvents.
• Hydrogen bonding interaction between solute and solvent may lead to high solubility. (In alcohols as the length of the carbon chain increase, the solubility of alcohols in water decreases)
• The solubility of ionic compounds in water varies from one solid to another.(strength of bonding forces within the solid)
• - Nonpolar substances are soluble in nonpolar substances.
Question: predict whether each of the following subtances is more likely to dissolve in CCl4 or in H2O

C7H16 NaHCO3 HCl I2

Solubility & Temperature
• Solubility of most of solids increase with temperature.
• Solubility of gases decrease with increased temperature

What mass of NH4Cl will crystallize when the solution is cooled to 20 °C?

The solubility of NH4Cl

At 20 °C is 37gNH4Cl/100gH2O

At 60 °C is 56gNH4Cl/100gH2O

Solubility & Pressure

• Solubility of a gas increases with increasing pressure

Henry’sLaw

C = k Pgas

C = solubility of gas

k = Henry’sLaw constant

Pgas = is the partial pressure of gas above the solution

The solubility of pure nitrogen in blood at body temperature (37°C) and 1 atm is 6.2x10^-4M. If a diver breathes air (XN2= 0.78) at a depth where the total pressure is 2.5atm, calculate the concentration of nitrogen in his blood?

Vapor Pressures of Solutions
• Roault, 1880s.
• Dissolved solute lowers vapor pressure of solvent.
• The partial pressure exerted by solvent vapor above an ideal solution is the product of the mole fraction of solvent in the solution and the vapor pressure of the pure solvent at a given temperature.

PA = A P°

A= Mole fraction

P° = vapor pressure of pure solvent

Calculate the vapor pressure lowering caused by the addition of 100g sucrose (C12H22O11) to 1000g of water. (water pressure of pure water at 25°C is 23.8mmHg.
A solution contains 102g of sugar (Mw=342.30g/mol) in 375g of water. Calculate the vapor pressure lowering at 25°C (vp of pure water 23.76mmHg at 25°C )
The vapor pressures of pure benzene and pure toluen at 25°C are 95.1 and 28.4mmHg respectively,when 1 mol benzene mixed with 1 mol toluen calculate the total vapor presure of the solution.

When a solution and a pure solvent (or two solution of different concentration) are separated by a semipermeable membrane, solvent molecules pass through the membrane in a process called osmosis.

n

π = RT

V

Osmotic Pressure

When a solution and a pure solvent (or two solution of different concentration) are separated by a semipermeable membrane, solvent molecules pass through the membrane in a process called osmosis.

The osmotic pressure is equal to the external pressure to prevent osmosis

πV = nRT

= M RT

For dilute solutions of electrolytes:

The average osmotic pressure of blood is 7.7atm at 25°C. What concentration of glucose will be isotonic with blood.
• İsotonic solutions are the solutions which have same osmotic pressure.
Freezing-Point Depression and Boiling Point Elevation

ΔTb = -Kbi m

ΔTf = -Kfi m

m= molality i= van’t Hoff factor

van’t Hoff

measured ΔTf

0.0361°C

i = = = 1.98

expected ΔTf

0.0186°C

π = -i M RT

ΔTf = -i Kfm

ΔTb = -i Kb m

List the following solutions in order to their expected freezing points
• 0.050m CaCl2
• 0.150m NaCl
• 0.100m HCl
• 0.050m HC2H3O2 (acetic acid)
• 0.100m C12H22O11
Calculate the freezing point and the boiling point of a solution of 100.0g ethyleneglicol (C2H6O2) in 900.0g H2O
• For water Kb=0.52°C/m Kf= 1.86 °C/m
The osmotic pressure of 5g/L hemoglobin solution at 25°C is 0.0018atm, calculate the Mw of hemoglobin
When 0.2 of a nonionic subtances dissolved in 50g puer water the freezing point of this solution is found -0.22 °C, calculate the Mw of the subtances
To determine the molar mass of vasopressin, 10.00g of vasopressin is dissolved in 50.00g of Naphthalene (kf=6.94°C/m) the freezing point of the mixture is determined to be 79.01°C, that of pure Naphthalene is 80.29°C. what is the molar mass of vasopressin?

Naphthalene (Mw=128.2g/mol) is the active ingredient of moth balls. in a solution prepared by mixing 25.0g Naphthalene with 0.750L of CS2 (d=1.263g/ml). Assume that volume remains 0.750L when the solution is prepared.

• what is the mass percent of Naphthalene in the solution
• what is the concentration of Naphthalene in ppm
• what is the density of the solution
• what is the molarity of Naphthalene in the solution
• what is the molality of Naphthalene in the solution
• the vapor pressure of pure CS2 at 25°C is 358mmHg. Assume that the vapor pressure exerted by Naphthalene at 25°C is negligible. What is the vapor pressure of the solution at this temperature?
• What is the osmotic pressure of the solution
• The normal boiling point of CS2 is 46.13°C (Kb= 2.34°C/m) what is the normal boiling point of the solution?