Question

Chloroform is a common liquid used in the laboratory. It vaporizes readily. If the pressure of chloroform vapori in a flask is \(195 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C}\) and the density of the vapor is \(1.25 \mathrm{g} / \mathrm{L},\) what is the molar mass of chloroform?

Short answer

The molar mass of chloroform is approximately 119.13 g/mol.

Step-by-step solution

Convert Pressure to atm

The given vapor pressure of chloroform is 195 mm Hg. To use this in calculations, convert it to atm: \[ \text{Pressure in atm} = \frac{195 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 0.2566 \text{ atm} \]

Apply Ideal Gas Law

Use the Ideal Gas Law equation, \( PV = nRT \), to find the molar mass.1. Rearrange to find the volume per mole (\(V/n\)): \[ V/n = \frac{RT}{P} \]2. At 25.0°C (which is 298 K) and with R = 0.0821 L·atm/K·mol, plug in the values: \[ \frac{V}{n} = \frac{0.0821 \times 298}{0.2566} \approx 95.3 \text{ L/mol} \]

Find Molar Mass From Density

Using the given density \( d = 1.25 \text{ g/L} \), calculate the molar mass \( M \):\[ \text{Molar Mass} = \text{Density} \times \frac{V}{n} \]\[ M = 1.25 \text{ g/L} \times 95.3 \text{ L/mol} \approx 119.125 \text{ g/mol} \]

Report Final Result

The calculated molar mass of chloroform is approximately \( 119.13 \text{ g/mol} \). This value helps verify the identity of the vaporized substance in the flask.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a crucial concept in chemistry, especially when dealing with gases. It's represented by the formula \( PV = nRT \), which relates four important properties of gases: pressure \( P \), volume \( V \), number of moles \( n \), and temperature \( T \). Here, \( R \) is the ideal gas constant, commonly used as \( 0.0821 \) L·atm/K·mol.

In the original exercise, the Ideal Gas Law was used to calculate the volume that one mole of the gas would occupy under the given conditions. By rearranging the equation to \( V/n = RT/P \), we can find this volume. This rearrangement is particularly useful when combined with other properties like density, allowing us to solve for molar mass.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase. For any given substance, it increases with temperature as more particles acquire enough energy to escape into the gas phase.

In this exercise, the vapor pressure of chloroform is given as 195 mm Hg. To use it in calculations with the Ideal Gas Law, it was converted to atmospheres (atm). This conversion is essential because the gas constant \( R \) in the Ideal Gas Law equation is typically given in atm. By knowing the vapor pressure, we can predict how readily a substance will evaporate.

Chemical Calculations

Chemical calculations often involve converting between different units and using equations to find unknowns. They can encompass the conversion of pressure units, rearrangement of equations, and combining properties like density to find molar mass.

In the original problem, chemical calculations were used to convert the vapor pressure from mm Hg to atm and to apply the Ideal Gas Law. These conversions and calculations help to manipulate and solve specific chemical contexts, bringing clarity to often complex situations.

Density and Volume Relationships

Density is defined as mass per unit volume, and in gases, it's used to determine molar mass when combined with volume data. A key relationship shown in the exercise is: the molar mass \( M \) can be calculated using \( M = \text{Density} \times V/n \), where \( V/n \) is the volume one mole of the gas occupies.

From the exercise, we know that chloroform has a vapor density of 1.25 g/L. Using the calculated molar volume, we find the molar mass of chloroform. This linkage between density and volume is a foundational concept to understand how different properties of a substance interact and can be used to derive one from another. Understanding how to manipulate these relationships provides powerful tools in solving a variety of chemical problems.