Part 2 Definitions Flashcards
(34 cards)
What is molarity and molality?
Molarity in mol/L
amount solute (in mol) / volume solution (in L)
Molality (m) in mol/kg
amount solute (in mol) / mass solvent (in kg)
What is mole fraction and mole percent?
mole fraction (X) with no units
amount solute (in mol) / total amount of solution and solvent (in mol)
mole percent (mol %) with % units
(amount solute (in mol) / total amount of solution and solvent (in mol)) x 100%
What is percent by mass
in parts per million by mass (ppm)
and in parts per billion (ppb)
What is parts by volume (%, ppm, ppb)?
volume solute / volume solution x multiplication factor
multiplication factor can be %, ppm, ppb
what are the multiplication factors for solubility concentration terms?
What can the multiplication factor be used for in the solubility concentration terms?
percent: multiplication is 100%
parts per million (ppm): multiplication factor is 10^6
parts per billion (ppb): multiplication factor is 10^9
Multiplication factors can be used in percent/ppm/ppb by mass
OR parts by volume in %, ppm, ppb
Which is temperature dependent molarity or molality?
Molarity is temp dependent
Molality IS NOT temp dependent
What are some tips for colligative properties problems?
- Always start writing what the problem gives you
- If we know the density of the solution, we can calculate the molality from the molarity and vice versa.
o Molarity is temp dependent
o Molality IS NOT temp dependent
• Don’t get frazzled, cancel your units!
An aqueous nitric acid is 68% by mass and has a density of 1.40 g/mL
Calculate molarity, molality, mole fraction
Make assumption we have 100g of solution of HNO3
68g of HNO3 (from 68% by mass)
32g of Water (must be 32% mass remaining)
Mass solution = mass solvent + mass solute = 100g
Molality = moles of solute/kg of solvent
Moles of solute = 68g HNO3 x (1mol/63.01g) = 1.08 mol HNO3
Kg of solvent = 32g x (1 kg/1000g) = 0.032 kg H20
Molality = (1.08mol H20/0.032 kg H20)
Molality (m) = 33.7 molal or mol/kg
Mole fraction = mol solute/mol solute + mol solvent
32g H20 x (1mol/18g H20) = 1.78 mol H20
From molality we know mol of HNO3 = 1.08mol
Mole fraction = 1.06 mol/1.08mol + 1.78mol
Mf = 0.38 (NO UNITS! Or little x after)
Molarity = mol solute/vol in L
To find Volume use the Density = m/v = 1.41 g/mL
From previous assumption the solution is 100g
100g (1 mL/1.40 g) = 74.1 mL solution
74.1 mL x (1 L/1000 mL) = 0.0741 L solution
Molarity (M) = (1.08mol HNO3/ 0.0714L) = 15.1 M
When you see a “x% by mass” what do you assume?
Make assumption we have 100g of solution of solvent
so the x% mass is of 100g
Determine the molarity of a solution prepared by dissolving 280.0 mg of NaCl in water to form 2.00 ml of solution
Determine the molality of a solution in which 12.9g fructose (CgH12Og) is dissolved in 31.0 g water
Determine the molarity of a solution prepared by dissolving 280.0 mg of NaCl in water to form 2.00 ml of solution
- 0mg x (10^3mg/1g) = 0.2800 g
- 00 ml x (1000 mL/1L) = 0.002 L solution
- 2800 g Nacl x (1 mol/58g NacL) = 0.00428
- 00428 mol NaCl / 0.002L solution =
Determine the molality of a solution in which 12.9g fructose (CgH12Og) is dissolved in 31.0 g water
Molality (m) = mol solute/kg solvent
12.9 g x (1 mol/180g fructose) = 0.0717 mol/0.031kg = 2.31m
What are colligative properties?
Property that depends on the number of solute particles, not their identity.
1) Vapor pressure lowering
2) Freezing Point Depression (Δtf)
3) Boiling Point Elevation (Δtb)
4) Osmotic Pressure
What happens when the vapor pressure lowers?
Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.
When we add solute, less molecule of solvent are available to escape ==> lower vapor pressure
Molecules at surface of liquid feel a different interaction than molecules in the bulk of the solution
The right energy is needed to escape. Equilibrium = vapor pressure
Pure solvent - only pure water molecules - only account for water-water interaction
Adding a nonvolatile solute (wont evaporate easily) water molecules interacting with something else and are less available to escape ==> less in gas phase ==> lower vapor pressure
What is Raoult’s Law?
What is it used for?
What can this be used for?
• Used to calculate the new vapor pressure (of the solution)
Psolution = Xsolvent x Psolvent
- Notice that, the molar fraction is telling us that we only care about the solvent.
- DON’T GET CONFUSED! We’re used to using molar fraction for the solute, not the solvent.
- The pressure only cares about the molecules of solvent (since those are the ones that can escape)
THIS IS ONLY FOR NONVOLATILE SOLUTES AND NONELECTROLYTES
From Raoult’s Law how do we determine CHANGE in vapor pressure (deltaP)
deltaP = Psolvent - Psolution
THIS IS ONLY FOR NONVOLATILE SOLUTES AND NONELECTROLYTES
The vapor pressure of water at 25 degrees Celsius is 23.8 mm Hg. What is the vapor pressure of a solution containing 5.50g of non-electrolyte sucrose (molar mass = 342 g/mol) in 12.8g water (molar mass = 18.0g/mol) at 24 degrees Celsius?
- 8 mmHg = P solvent
- 5g x (1mol/342g) = 0.0161 mol sucrose - SOLUTE
- 8g x (1 mol/18g) = 0.711 mol H20 - SOLVENT
X solvent = (0.711 mol H20/0.711 mol H20 + 0.0161 mol sucrose)
X solvent = 0.978
P solution = (0.978 x 23.8 mmHg) = 23.3 mmHg
deltaP = (23.8 mmHg - 23.3 mmHg) = 0.5 mmHg
what is vapor pressure in volatile solutes?
Now, both the solvent and the solute will escape to gas!
what is Ptotal?
P total = Pa + Pb
The total pressure is the sum of the partial pressures, which uses mole fractions
Pa = Xa * Pa
Pb = Xb * Pb
Partial pressure are just the true pressure times the molar fraction
THIS IS ONLY FOR ELECTROLYTES
At 20 degrees Celcius vapor pressures of pure benzene (C6H6 molar mass = 78.0 g/mol) and toluene (C6H5CH3, molar mass = 92.0 g/mol) are 22 mmHg and 75 mm Hg respectively. What is the total vapor pressure above a solution containing 20.0g of benzene and 20.0g of toluene at 20 degrees Celcius?
20g Benzene x (1 mol/78.0 g) = 0.256 mol Benzene
20g Toluene x (1 mol/92g) = 0.217 mol Toluene
Molar Fraction Benzene = 0.256 mol Benzene / 0.256 mol Benzene + 0.217 mol Toluene = 0.541xb
Molar Fraction Toluene = 0.217 mol Toluene / 0.256 mol Benzene + 0.217 mol Toluene = 0.459xt
Psolution = (xb * Pb) + (xt * Pt)
Psolution = (0.541) x (22 mmHg) + (0.459) x (74 mm Hg)
= 11.91 mmHg + 33.97 mmHg = 45.9 mmHg
What is freezing point depression?
Freezing point depression is the phenomena that describes why adding a solute to a solvent results in the lowering of the freezing point of the solvent. When a substance starts to freeze, the molecules slow down due to the decreases in temperature, and the intermolecular forces start to take over.
How does vapor pressure relate to lower melting point and higher boiling point?
- Remember: Because solute- solvent intermolecular attractions make it harder for solvent to escape to the vapor phase.
- Instead of fighting only solvent- solvent interactions, now we’re fighting solvent-solute interactions, which means we need more energy.
- That higher energy requirements reflects itself as freezing point depression and boiling point elevation.
Notice that the vapor pressure for the solution is shifted downward and to the left compared to that of the pure solvent. Consequently, the vapor pressure curve intersects the solid–gas curve at a lower temperature. The net effect is that the solution has a lower melting point and a higher boiling point than the pure solvent.
The freezing point of a solution containing a nonvolatile solute is lower than the freezing point of the pure solvent.
The more concentrated the solution, the lower the freezing point becomes.
what is boiling point of elevation?
Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. This happens whenever a non-volatile solute, such as a salt, is added to a pure solvent, such as water.
