Freezing Point Depression Of Nacl
xiii.8: Freezing-Point Depression and Boiling-Point Elevation of Nonelectrolyte Solutions
- Folio ID
- 24260
Learning Objectives
- To depict the relationship between solute concentration and the concrete properties of a solution.
- To sympathise that the full number of nonvolatile solute particles determines the decrease in vapor force per unit area, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.
Many of the physical properties of solutions differ significantly from those of the pure substances discussed in before chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its utilize. Aqueous solutions take both a lower freezing point and a higher humid indicate than pure water. Probably one of the nigh familiar applications of this phenomenon is the improver of ethylene glycol ("antifreeze") to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very common cold atmospheric condition from the expansion of pure h2o on freezing. Antifreeze also enables the cooling organization to operate at temperatures greater than 100°C without generating plenty pressure level to explode.
Changes in the freezing point and boiling bespeak of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are chosen colligative properties (from the Latin colligatus, meaning "bound together" as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties.
When we make up one's mind the number of particles in a solution, information technology is important to remember that non all solutions with the same molarity comprise the aforementioned concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, \(NaCl\), and \(\ce{CaCl_2}\). Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In dissimilarity, both \(\ce{NaCl}\) and \(\ce{CaCl_2}\) are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of \(\ce{NaCl}\) contains 0.01 M Na+ ions and 0.01 M \(Cl^−\) ions, for a total particle concentration of 0.02 M. Similarly, the \(\ce{CaCl_2}\) solution contains 0.01 M \(Ca^{two+}\) ions and 0.02 M \(Cl^−\) ions, for a full particle concentration of 0.03 M.These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At higher concentrations (typically >1 Thou), especially with salts of small-scale, highly charged ions (such as \(Mg^{ii+}\) or \(Al^{3+}\)), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the concrete properties of a solution. In the following discussion, nosotros must therefore proceed the chemical nature of the solute firmly in mind.
Humid Signal Elevation
Recall that the normal boiling betoken of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, information technology must also touch the boiling point. Because the vapor force per unit area of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal humid signal of the solvent. Thus the humid point of a solution is ever greater than that of the pure solvent. We tin encounter why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure \(\PageIndex{1}\)). The vapor force per unit area of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor bend for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water.
The boiling point of a solution with a nonvolatile solute is e'er greater than the boiling bespeak of the pure solvent.
The magnitude of the increase in the humid point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling bespeak must besides be proportional to the concentration of the solute (Effigy \(\PageIndex{two}\)). Nosotros tin define the humid betoken height (\(ΔT_b\)) as the difference between the boiling points of the solution and the pure solvent:
\[ΔT_b=T_b−T^0_b \label{eq1}\]
where \(T_b\) is the boiling point of the solution and \(T^0_b\) is the boiling indicate of the pure solvent. We can express the human relationship between \(ΔT_b\) and concentration equally follows
\[ΔT_b = mK_b \label{eq2}\]
where chiliad is the concentration of the solute expressed in molality, and \(K_b\) is the molal boiling point pinnacle abiding of the solvent, which has units of °C/m. Table \(\PageIndex{1}\) lists characteristic Kb values for several ordinarily used solvents.
For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration.
| Solvent | Humid Bespeak (°C) | Yardb (°C/1000) | Freezing Indicate (°C) | Kf (°C/m) |
|---|---|---|---|---|
| acerb acid | 117.ninety | iii.22 | 16.64 | 3.63 |
| benzene | eighty.09 | 2.64 | 5.49 | 5.07 |
| d-(+)-camphor | 207.4 | 4.91 | 178.8 | 37.viii |
| carbon disulfide | 46.2 | 2.42 | −112.1 | 3.74 |
| carbon tetrachloride | 76.eight | five.26 | −22.62 | 31.4 |
| chloroform | 61.17 | 3.eighty | −63.41 | 4.60 |
| nitrobenzene | 210.viii | 5.24 | 5.70 | half-dozen.87 |
| h2o | 100.00 | 0.51 | 0.00 | 1.86 |
The concentration of the solute is typically expressed equally molality rather than mole fraction or molarity for ii reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between naught and one). Using molality allows u.s. to eliminate nonsignificant zeros.
Co-ordinate to Table \(\PageIndex{1}\), the molal humid point elevation constant for h2o is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at one.00 atm. The increase in the humid betoken of a 1.00 m aqueous \(\ce{NaCl}\) solution volition be approximately twice equally large as that of the glucose or sucrose solution because 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved ions. Hence a 1.00 one thousand \(\ce{NaCl}\) solution will have a humid point of about 101.02°C.
Example \(\PageIndex{3}\)
In Example \(\PageIndex{1}\), we calculated that the vapor pressure of a thirty.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor force per unit area of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution.
Given: composition of solution
Asked for: humid point
Strategy:
Summate the molality of ethylene glycol in the 30.ii% solution. Then utilize Equation \ref{eq2} to summate the increase in humid point.
Solution:
From Example \(\PageIndex{1}\), we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 1000 of h2o. The molality of the solution is thus
\[\text{molality of ethylene glycol}= \left(\dfrac{iv.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \correct)=6.98 chiliad\]
From Equation \ref{eq2}, the increment in boiling bespeak is therefore
\[ΔT_b=mK_b=(half-dozen.98 \cancel{g})(0.51°C/\cancel{m})=3.half dozen°C\]
The boiling indicate of the solution is thus predicted to exist 104°C. With a solute concentration of most 7 k, however, the assumption of a dilute solution used to obtain Equation \ref{eq2} may not be valid.
Exercise \(\PageIndex{3}\)
Presume that a tablespoon (5.00 m) of \(\ce{NaCl}\) is added to 2.00 L of water at xx.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water eddy?
- Respond
-
100.04°C, or 100°C to three significant figures. (Call back that 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved particles. The small increment in temperature ways that calculation table salt to the water used to cook pasta has substantially no consequence on the cooking time.)
Freezing Signal Depression
The phase diagram in Figure \(\PageIndex{ane}\) shows that dissolving a nonvolatile solute in water not simply raises the boiling point of the h2o only as well lowers its freezing indicate. The solid–liquid curve for the solution crosses the line corresponding to P = 1 atm at a lower temperature than the curve for pure h2o.
This phenomenon is exploited in "de-icing" schemes that use salt (Figure \(\PageIndex{3}\)), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an "antifreeze" in automobile radiators. Seawater freezes at a lower temperature than fresh water, and and then the Chill and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the torso fluids of fish and other cold-blooded sea animals that alive in these oceans).
We can understand this event by imagining that nosotros have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. H2o molecules are continuously colliding with the ice surface and entering the solid phase at the same charge per unit that water molecules are leaving the surface of the ice and entering the liquid phase. If we deliquesce a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules volition reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the water ice lattice. Consequently, the presence of glucose molecules in the solution can simply subtract the rate at which water molecules in the liquid collide with the water ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to cook. The only manner to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the charge per unit at which water molecules exit the surface of the ice crystals until information technology equals the charge per unit at which h2o molecules in the solution collide with the ice.
By analogy to our treatment of boiling point elevation,the freezing point depression (\(ΔT_f\)) is divers equally the difference between the freezing point of the pure solvent and the freezing indicate of the solution:
\[ ΔT_f=T^0_f−T_f \label{eq3}\]
where
- \(T^0_f\) is the freezing point of the pure solvent and
- \(T_f\) is the freezing point of the solution.
The club of the terms is reversed compared with Equation \ref{eq1} to express the freezing signal depression as a positive number. The relationship between \(ΔT_f\) and the solute concentration is given by an equation coordinating to Equation \ref{eq2}:
\[ΔT_f = mK_f \label{eq4}\]
where
- \(m\) is the molality of the solution and
- \(K_f\) is the molal freezing point depression abiding for the solvent (in units of °C/chiliad).
Like \(K_b\), each solvent has a characteristic value of \(K_f\) (Table \(\PageIndex{1}\)). Freezing betoken depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point pinnacle. Thus an aqueous \(\ce{NaCl}\) solution has twice as large a freezing point depression as a glucose solution of the same molality.
People who live in cold climates utilise freezing point low to their advantage in many ways. For instance, ethylene glycol is added to engine coolant water to prevent an automobile engine from beingness destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and forbid the formation of more ice, which would be particularly unsafe if formed on the command surfaces of the aircraft (Video \(\PageIndex{1}\)).
Video \(\PageIndex{1}\): Freezing point depression is exploited to remove ice from the command surfaces of aircraft.
The decrease in vapor pressure level, increment in boiling point, and subtract in freezing signal of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles.
Example \(\PageIndex{4}\)
In colder regions of the U.s.a., \(\ce{NaCl}\) or \(\ce{CaCl_2}\) is often sprinkled on icy roads in winter to cook the ice and make driving safer. Utilise the data in Figure xiii.9 to estimate the concentrations of two saturated solutions at 0°C, one of \(\ce{NaCl}\) and i of \(\ce{CaCl_2}\), and summate the freezing points of both solutions to see which salt is likely to be more constructive at melting ice.
Given: solubilities of 2 compounds
Asked for: concentrations and freezing points
Strategy:
- Estimate the solubility of each table salt in 100 g of water from Figure xiii.ix. Determine the number of moles of each in 100 g and calculate the molalities.
- Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation \(\PageIndex{4}\) to calculate the freezing point depressions of the solutions.
Solution:
A From Figure 13.nine, we can gauge the solubilities of \(\ce{NaCl}\) and \(\ce{CaCl_2}\) to be about 36 g and 60 thousand, respectively, per 100 thou of water at 0°C. The corresponding concentrations in molality are
\[m_{\ce{NaCl}}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\abolish{yard} \;H_2O}\correct)\left(\dfrac{ane\; mol\; NaCl}{58.44\; \abolish{ grand\; NaCl}}\right)\left(\dfrac{1000\; \cancel{chiliad}}{1\; kg}\right)=6.ii\; k\]
\[m_{\ce{CaCl_2}}=\left(\dfrac{threescore\; \cancel{grand\; CaCl_2}}{100\;\abolish{thou}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{grand \;\abolish{g}}{one kg}\correct)=5.4\; grand\]
The lower formula mass of \(\ce{NaCl}\) more than compensates for its lower solubility, resulting in a saturated solution that has a slightly college concentration than \(\ce{CaCl_2}\).
B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of measurement of \(\ce{NaCl}\) and \(\ce{CaCl_2}\), respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 thousand = 12 thou for \(\ce{NaCl}\) and 3 × 5.iv chiliad = 16 m for \(\ce{CaCl_2}\). The resulting freezing indicate depressions can exist calculated using Equation \(\PageIndex{4}\):
\[\ce{NaCl}: ΔT_f=mK_f=(12\; \abolish{m})(1.86°C/\cancel{m})=22°C\]
\[\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{thou})(i.86°C/\abolish{k})=30°C\]
Because the freezing indicate of pure h2o is 0°C, the actual freezing points of the solutions are −22°C and −thirty°C, respectively. Annotation that \(\ce{CaCl_2}\) is essentially more effective at lowering the freezing bespeak of water because its solutions comprise three ions per formula unit. In fact, \(\ce{CaCl_2}\) is the salt usually sold for home use, and information technology is also oftentimes used on highways.
Considering the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by just a sure amount, regardless of how much common salt is spread on an icy route. If the temperature is significantly below the minimum temperature at which one of these salts volition cause ice to melt (say −35°C), in that location is no bespeak in using common salt until it gets warmer
Exercise \(\PageIndex{iv}\)
Calculate the freezing betoken of the thirty.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Example \(\PageIndex{half dozen}\).viii and Case \(\PageIndex{6}\).10.
- Reply
-
−xiii.0°C
Example \(\PageIndex{5}\)
Accommodate these aqueous solutions in social club of decreasing freezing points: 0.1 m \(KCl\), 0.1 k glucose, 0.one m SrCl2, 0.i k ethylene glycol, 0.1 m benzoic acid, and 0.one m HCl.
Given: molalities of six solutions
Asked for: relative freezing points
Strategy:
- Place each solute as a strong, weak, or nonelectrolyte, and employ this information to decide the number of solute particles produced.
- Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point low.
Solution:
A Because the molal concentrations of all 6 solutions are the aforementioned, we must focus on which of the substances are stiff electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. \(KCl\), \(SrCl_2\), and \(HCl\) are stiff electrolytes, producing two, 3, and two ions per formula unit of measurement, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (i particle per molecule).
B The molalities of the solutions in terms of the total particles of solute are: \(KCl\) and \(HCl\), 0.2 m; \(SrCl_2\), 0.3 m; glucose and ethylene glycol, 0.ane 1000; and benzoic acrid, 0.one–0.two m. Because the magnitude of the subtract in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing bespeak depression) > benzoic acid > \(HCl\) = \(KCl\) > \(SrCl_2\).
Practise \(\PageIndex{5}\)
Arrange these aqueous solutions in order of increasing freezing points: 0.2 thousand \(NaCl\), 0.3 m acetic acid, 0.1 thou \(\ce{CaCl_2}\), and 0.2 thou sucrose.
- Answer
-
0.2 m \(\ce{NaCl}\) (everyman freezing point) < 0.3 m acetic acid ≈ 0.1 chiliad \(\ce{CaCl_2}\) < 0.two m sucrose (highest freezing point)
Boiling Signal Elevation and Freezing Betoken Low: https://youtu.be/0MZm1Ay6LhU
Determination of Molar Masses
Osmotic force per unit area and changes in freezing signal, boiling point, and vapor pressure level are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.
Determining Molar Mass from Freezing Point Depression
A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?
Solution
We tin solve this problem using the following steps.
- Determine the alter in freezing point from the observed freezing indicate and the freezing point of pure benzene (Table \(\PageIndex{ane}\)).
\(ΔT_\ce{f}=\mathrm{v.five\:°C−ii.32\:°C=3.2\:°C}\)
- Determine the molal concentration from Chiliad f, the freezing signal low constant for benzene (Table \(\PageIndex{1}\)), and ΔT f.
\(ΔT_\ce{f}=K_\ce{f}m\)
\(thou=\dfrac{ΔT_\ce{f}}{K_\ce{f}}=\dfrac{3.2\:°\ce C}{5.12\:°\ce C m^{−1}}=0.63\:g\)
- Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
\(\mathrm{Moles\: of\: solute=\dfrac{0.62\:mol\: solute}{1.00\cancel{kg\: solvent}}×0.0550\cancel{kg\: solvent}=0.035\:mol}\)
- Make up one's mind the molar mass from the mass of the solute and the number of moles in that mass.
\(\mathrm{Molar\: mass=\dfrac{four.00\:grand}{0.034\:mol}=1.2×x^2\:g/mol}\)
Exercise \(\PageIndex{6}\)
A solution of 35.seven g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this chemical compound?
- Answer
-
1.8 × 102 g/mol
Conclusion of a Molar Mass from Osmotic Pressure
A 0.500 L sample of an aqueous solution containing x.0 chiliad of hemoglobin has an osmotic force per unit area of five.9 torr at 22 °C. What is the molar mass of hemoglobin?
Solution
Here is one gear up of steps that can exist used to solve the problem:
- Convert the osmotic pressure to atmospheres, and then determine the molar concentration from the osmotic pressure level.
\[\Pi=\mathrm{\dfrac{5.9\:torr×i\:atm}{760\:torr}=seven.8×x^{−3}\:atm}\]
\[\Pi=MRT\]
\(Grand=\dfrac{Π}{RT}=\mathrm{\dfrac{7.8×10^{−3}\:atm}{(0.08206\:50\: atm/mol\: Thou)(295\:K)}=3.ii×ten^{−four}\:Thou}\)
- Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.
\(\mathrm{moles\: of\: hemoglobin=\dfrac{iii.2×ten^{−4}\:mol}{1\cancel{Fifty\: solution}}×0.500\cancel{L\: solution}=1.six×ten^{−4}\:mol}\)
- Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
\(\mathrm{molar\: mass=\dfrac{x.0\:g}{1.six×10^{−four}\:mol}=6.2×10^4\:one thousand/mol}\)
Exercise \(\PageIndex{7}\)
What is the molar mass of a poly peptide if a solution of 0.02 g of the poly peptide in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?
- Answer
-
ii.seven × 104 one thousand/mol
Finding the Molecular Weight of an Unknown using Colligative Properties:
https://youtu.be/faSk2REYy74
Summary
- vapor pressure lowering: \[P^0_A−P_A=ΔP_A=X_BP^0_A \]
- vapor force per unit area of a system containing 2 volatile components: \[P_T=X_AP^0_A+(1−X_A)P^0_B \]
- boiling point elevation: \[ΔT_b = mK_b\]
- freezing signal depression: \[ΔT_f = mK_f \]
The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor force per unit area, boiling point, freezing indicate, and osmotic force per unit area. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known equally Raoult's law. Solutions that obey Raoult's law are called platonic solutions. Most real solutions exhibit positive or negative deviations from Raoult'southward police force. The boiling point elevation (\(ΔT_b\)) and freezing bespeak depression (\(ΔT_f\)) of a solution are defined as the differences betwixt the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute.
Freezing Point Depression Of Nacl,
Source: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.08%3A_Freezing-Point_Depression_and_Boiling-Point_Elevation_of_Nonelectrolyte_Solutions
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