Problem 100
Question
A solution contains \(0.115 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(30^{\circ} \mathrm{C}\) is \(25.7\) torr. The vapor pressure of pure water at this temperature is \(31.8\) torr. Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)
Step-by-Step Solution
Verified Answer
There are approximately 0.826 grams of sodium chloride in the solution.
1Step 1: Write down the expression for Raoult's law
Raoult's Law states that the partial pressure of a component in a solution is proportional to its mole fraction. For a solution of water and sodium chloride, the relationship between the vapor pressure of the solution (P_solution) and the mole fraction of water (X_H2O) can be expressed as:
P_solution = X_H2O × P_pure_water
where P_pure_water is the vapor pressure of pure water.
2Step 2: Calculate the mole fraction of water
To find the mole fraction of water, we need to determine the total moles of particles in the solution. Since sodium chloride is a strong electrolyte, it will dissociate completely into its ions (Na+ and Cl-) upon dissolving in water. We can represent the dissociation of sodium chloride as follows:
\(NaCl_{(aq)} \longrightarrow Na^+_{(aq)} + Cl^-_{(aq)}\)
Let x be the unknown number of moles of NaCl. Its dissociation into its two ions doubles the contribution of particles in the solution, making it 2x ions. The mole fraction of water can then be expressed as:
\( X_{H_{2}O} = \frac {0.115 \mathrm{~moles} \mathrm{~H_{2}O}}{0.115\mathrm{~moles} \mathrm{~H_{2}O}+ 2\mathrm{~x~moles}\mathrm{~NaCl}}\)
3Step 3: Solve for the unknown number of moles of sodium chloride
Substitute the given values of P_solution, P_pure_water, and X_H2O in the Raoult's Law equation:
\(25.7\,\mathrm{torr} = \frac {0.115\,\mathrm{~mol}\,\mathrm{H}_{2} \mathrm{O}}{0.115\,\mathrm{~mol}\,\mathrm{H}_{2} \mathrm{O}+ 2\,\mathrm{x~mol}\,\mathrm{NaCl}} \times 31.8\, \mathrm{torr}\)
Now solve for x:
\(x = \frac {31.8\,\mathrm{torr} - 25.7\,\mathrm{torr}} {25.7\, \mathrm{torr} \times 2} \times 0.115\, \mathrm{mol}\, \mathrm{H}_{2} \mathrm{O}\)
\(x ≈ 0.01415\, \mathrm{mol}\, \mathrm{NaCl}\)
4Step 4: Calculate the grams of sodium chloride
To find the grams of sodium chloride, multiply the number of moles by the molar mass of NaCl:
Mass of NaCl = moles of NaCl × molar mass of NaCl
The molar mass of sodium chloride is 58.44 g/mol. Therefore,
Mass of NaCl = 0.01415 mol × 58.44 g/mol ≈ 0.826 g
So, there are approximately 0.826 grams of sodium chloride in the solution.
Key Concepts
Mole Fraction CalculationColligative PropertiesSodium Chloride Dissociation
Mole Fraction Calculation
When working with solutions in chemistry, understanding the mole fraction is crucial for many calculations, including the application of Raoult's Law. The mole fraction is defined as the ratio of the number of moles of a particular component to the total number of moles of all components in the mixture.
For instance, if a solution contains water and another substance, like sodium chloride, the mole fraction of water (\( X_{H_2O} \)) would be calculated by dividing the moles of water by the sum of moles of water and the effective number of moles of sodium chloride, taking into account its dissociation into ions. The equation for the mole fraction of water in such a solution is:
\[ X_{H_{2}O} = \frac {\text{moles of } H_{2}O}{\text{moles of } H_{2}O + \text{moles of dissociated NaCl (ions)}} \]
Remember that for ionic compounds like sodium chloride, which dissociate in solutions, the number of particles increases as each NaCl molecule becomes two separate ions (Na\textsuperscript{+} and Cl\textsuperscript{-}). Therefore, in mole fraction calculations for such solutions, it’s critical to account for this increase in the number of particles. It directly influences the vapor pressure of the solution as described by Raoult's Law.
For instance, if a solution contains water and another substance, like sodium chloride, the mole fraction of water (\( X_{H_2O} \)) would be calculated by dividing the moles of water by the sum of moles of water and the effective number of moles of sodium chloride, taking into account its dissociation into ions. The equation for the mole fraction of water in such a solution is:
\[ X_{H_{2}O} = \frac {\text{moles of } H_{2}O}{\text{moles of } H_{2}O + \text{moles of dissociated NaCl (ions)}} \]
Remember that for ionic compounds like sodium chloride, which dissociate in solutions, the number of particles increases as each NaCl molecule becomes two separate ions (Na\textsuperscript{+} and Cl\textsuperscript{-}). Therefore, in mole fraction calculations for such solutions, it’s critical to account for this increase in the number of particles. It directly influences the vapor pressure of the solution as described by Raoult's Law.
Colligative Properties
Colligative properties are physical properties of solutions that depend on the number of dissolved particles in the solution, rather than the identity of those particles. Examples of colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
By applying Raoult's Law, which is a fundamental equation for colligative properties, we can relate the drop in vapor pressure of the solution to the presence of a non-volatile solute, such as sodium chloride. According to this law, a solution's vapor pressure decreases when a solute is added, because the solute molecules occupy space at the surface of the liquid and reduce the number of solvent molecules that can escape. This principle is used to calculate the mole fraction of the solute, where a lower vapor pressure indicates a higher proportion of solute particles.
Colligative properties are vital in many industries and can be observed in everyday phenomena, such as the decrease in freezing point of water with added salt, which is why salt is used to melt ice on roads in cold climates.
By applying Raoult's Law, which is a fundamental equation for colligative properties, we can relate the drop in vapor pressure of the solution to the presence of a non-volatile solute, such as sodium chloride. According to this law, a solution's vapor pressure decreases when a solute is added, because the solute molecules occupy space at the surface of the liquid and reduce the number of solvent molecules that can escape. This principle is used to calculate the mole fraction of the solute, where a lower vapor pressure indicates a higher proportion of solute particles.
Colligative properties are vital in many industries and can be observed in everyday phenomena, such as the decrease in freezing point of water with added salt, which is why salt is used to melt ice on roads in cold climates.
Sodium Chloride Dissociation
Sodium chloride (NaCl), commonly known as table salt, is an ionic compound that easily dissociates into its constituent ions, sodium (\textsuperscript{+}Na) and chloride (\textsuperscript{-}Cl), in water. This dissociative behavior has crucial implications in the realm of colligative properties and the application of Raoult's Law.
When NaCl dissolves in water:
\[ NaCl_{(aq)} \longrightarrow Na^+_{(aq)} + Cl^-_{(aq)} \]
Each unit of NaCl yields two separate ions – doubling the number of particles in the solution. Hence, when we consider mole fraction calculations for NaCl solutions, we must consider the increased count of solute particles to accurately determine colligative properties. Failure to account for the dissociation of NaCl into two ions would result in incorrectly calculating the solution's vapor pressure, boiling point, or freezing point.
In our exercise example, since sodium chloride is a strong electrolyte, for every mole of NaCl that dissociates, we must consider two moles of particles in the calculation for the mole fraction and in turn for the use in Raoult's Law, which is essential to solving problems in solution chemistry.
When NaCl dissolves in water:
\[ NaCl_{(aq)} \longrightarrow Na^+_{(aq)} + Cl^-_{(aq)} \]
Each unit of NaCl yields two separate ions – doubling the number of particles in the solution. Hence, when we consider mole fraction calculations for NaCl solutions, we must consider the increased count of solute particles to accurately determine colligative properties. Failure to account for the dissociation of NaCl into two ions would result in incorrectly calculating the solution's vapor pressure, boiling point, or freezing point.
In our exercise example, since sodium chloride is a strong electrolyte, for every mole of NaCl that dissociates, we must consider two moles of particles in the calculation for the mole fraction and in turn for the use in Raoult's Law, which is essential to solving problems in solution chemistry.
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