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What is the approximate boiling point at standard pressure of a solution prepared by dissolving \(234 \mathrm{~g}\) of \(\mathrm{NaCl}\) in \(500 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)

Short Answer

Expert verified
The approximate boiling point of the solution at standard pressure is \(104.110^{\circ}\mathrm{C}\).

Step by step solution

01

Calculate the number of moles of NaCl in the solution.

To calculate the number of moles of NaCl, use its molar mass, which is 58.44 g/mol. moles of NaCl = mass / molar mass = 234 g / 58.44 g/mol ≈ 4.006 mol
02

Calculate the molality of the solution.

Now that we have the number of moles of NaCl, we can find the molality of the solution. The mass of solvent (water) is given in grams, so we need to convert it to kilograms. mass of water = 500 g = 0.500 kg molality(m) = moles of solute / mass of solvent (in kg) = 4.006 mol / 0.500 kg ≈ 8.012 mol/kg
03

Calculate the boiling point elevation.

With the molality of the solution, we can now find the boiling point elevation using the boiling point elevation formula. ΔT = kB * m = 0.512 °C/m * 8.012 mol/kg ≈ 4.110 °C
04

Calculate the approximate boiling point of the solution.

Now that we have the boiling point elevation, we can add it to the normal boiling point of water to find the approximate boiling point of the solution. normal boiling point of water = 100 °C approximate boiling point of the solution = normal boiling point + boiling point elevation = 100 °C + 4.110 °C ≈ 104.110 °C So the approximate boiling point of the solution at standard pressure is 104.110 °C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molality
Molality, often denoted by the symbol 'm', is a measure of the concentration of a solute in a solution. Unlike molarity, which depends on the volume of the solution, molality is based solely on the mass of the solvent. It is defined as the number of moles of solute per kilogram of solvent.

To calculate the molality of a solution, you use the formula:
\[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\].

This calculation is especially useful when dealing with changes in temperature and pressure because, unlike liquid volume, the mass does not change with these conditions. In our textbook exercise, we used molality to determine how much the boiling point of water would elevate when a certain amount of sodium chloride (NaCl) was added.
Colligative Properties
Colligative properties are characteristics of solutions that depend on the number of particles in a solution but not on the type of particles. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.

Boiling point elevation, relevant to our exercise, is one of these properties and simply states that a solution has a higher boiling point than the pure solvent. This effect occurs because solute particles disrupt the solvent's ability to enter the gas phase, requiring more energy—thus a higher temperature—to boil. As such, the more particles (or moles) of solute you have in a solution, the greater the impact on the colligative properties.
Molar Mass
Molar mass is an essential concept in chemistry that relates the mass of a substance to its quantity in moles. It is defined as the mass of one mole of a given substance and is expressed in grams per mole (g/mol).

The molar mass of a compound like sodium chloride (NaCl) is the sum of the molar masses of its constituent elements, calculated from the periodic table. For NaCl, with sodium (Na) weighing approximately 22.99 g/mol and chlorine (Cl) approximately 35.45 g/mol, their combined molar mass is 58.44 g/mol. For our exercise, we used the molar mass of NaCl to convert grams of NaCl into moles, a necessary step to find the molality and subsequently the boiling point elevation.
Boiling Point Elevation Formula
The boiling point elevation formula tells us exactly how much the boiling point of a solvent will be raised when a solute is dissolved in it. The formula is given by:
\[\Delta T_b = i \cdot k_b \cdot m\]
where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van’t Hoff factor (which indicates the number of particles the solute splits into or forms in solution), \(k_b\) is the ebullioscopic constant of the solvent (a property specific to each solvent), and \(m\) is the molality of the solution.

In the steps provided, we calculated the boiling point elevation for our NaCl solution using water's ebullioscopic constant and the previously calculated molality. We then added this elevation to water’s normal boiling point to find the solution's boiling point. For many solvents, the ebullioscopic constant is a known value, and for water, it is 0.512 °C/m.

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Most popular questions from this chapter

A G.T.O. has a 22 gal. cooling system. Suppose you fill it with a \(50-50\) solution by volume of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\), ethylene glycol, and water. At what temperature would freezing become a problem? Assume the specific gravity of ethylene glycol is \(1.115\) and the freezing point depression constant of water is \(\left(1.86 \mathrm{C}^{\circ} / \mathrm{mole}\right)\). You might have placed in methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) instead of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\). If the current cost of ethylene glycol is 12 cents \(/ \mathrm{lb}\) and the cost of methanol is 8 cents \(/ \mathrm{lb}\), how much money would you save by using \(\mathrm{CH}_{3} \mathrm{OH}\) ? And yet, ethylene glycol is the more desirable antifreeze. Why? density \(=(.79 \mathrm{~g} / \mathrm{ml})\) for \(\mathrm{CH}_{3} \mathrm{OH}\) and \(3.785\) liter \(=1\) gallon.

You have two 1-liter containers connected to each other by a valve which is closed. In one container, you have liquid water in equilibrium with water vapor at \(25^{\circ} \mathrm{C}\). The other container contains a vacuum. Suddenly, you open the valve. Discuss the changes that take place, assuming temperature is constant with regard to (a) the vapor pressure, (b) the concentration of the water molecules in the vapor, (c) the number of molecules in the vapor state.

A chemist dissolves \(300 \mathrm{~g}\) of urea in \(1000 \mathrm{~g}\) of water. Urea is \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\). Assuming the solution obeys Raoult's'law, determine the following: a) The vapor pressure of the solvent at \(0^{\circ}\) and \(100^{\circ} \mathrm{C}\) and b) The boiling and freezing point of the solution. The vapor pressure of pure water is \(4.6 \mathrm{~mm}\) and \(760 \mathrm{~mm}\) at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), respectively and the \(\mathrm{K}_{\mathrm{f}}=(1.86\) \(\mathrm{C} / \mathrm{mole}\) ) and \(\mathrm{K}_{\mathrm{b}}=(52 \mathrm{C} / \mathrm{mole})\)

What percent of the AB particles are dissociated by water if the freezing point of \(a(.0100 \mathrm{~m})\) AB solution is \(-0.0193^{\circ} \mathrm{C}\) ? The freezing point lowering constant of water is \(\left(-1.86^{\circ} \mathrm{C} / \mathrm{mole}\right)\)

If the vapor pressure of \(\mathrm{CC} 1_{4}\) (carbon tetrachloride) is \(.132\) atm at \(23^{\circ} \mathrm{C}\) and \(.526 \mathrm{~atm}\) at \(58^{\circ} \mathrm{C}\), what is the \(\Delta \mathrm{H}^{\prime}\) in this temperature range?

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