Question

Chloroform, \(\mathrm{CHCl}_{3},\) is a volatile (easily vaporized) liquid solvent. Calculate the density of chloroform vapor at \(98^{\circ} \mathrm{C}\) and \(797 \mathrm{mmHg}\). Give the answer in grams per liter.

Short answer

The density of chloroform vapor at 98°C and 797 mmHg is approximately 4.05 g/L.

Step-by-step solution

Understand the Problem

We need to find the density of chloroform (\(\text{CHCl}_3\)) vapor at a specific temperature (98°C) and pressure (797 mmHg). To calculate density, we need the mass of chloroform in a given volume. We can use the ideal gas law to find the density.

Use Ideal Gas Law

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. Here, we will manipulate the equation to find density (\(d\)). First convert \(T\) to Kelvin, \(T = 98 + 273.15 = 371.15\,\text{K}\). The pressure in atm is \(P = \frac{797}{760}\, \text{atm} \approx 1.0487\, \text{atm}\). We will use \(R = 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\).

Calculate Molar Mass of CHCl3

Find the molar mass of chloroform: \(\text{C} = 12.01\, \text{g/mol}, \ \text{H} = 1.01\, \text{g/mol}, \ \text{Cl} = 35.45\, \text{g/mol}\). Thus, the molar mass of \(\text{CHCl}_3\) is \(12.01 + 1.01 + 3\times35.45 = 119.37\, \text{g/mol}.\)

Derive the Density Formula

Using the ideal gas law, rearrange to find density \(d = \frac{m}{V} = \frac{MP}{RT}\), where \(M\) is molar mass, \(P\) is pressure, \(R\) is the gas constant, and \(T\) is temperature. Use the values: \(M = 119.37 \ \text{g/mol}\), \(P = 1.0487 \ \text{atm}\), \(R = 0.0821 \ \text{L}\cdot\text{atm}/\text{K}\cdot\text{mol}\), \(T = 371.15 \ \text{K}\).

Calculate Density

Substitute the known values into the formula: \( d = \frac{119.37 \times 1.0487}{0.0821 \times 371.15}\). Calculate this to obtain \(d \approx 4.05\, \text{g/L}.\) Hence, the density of the chloroform vapor is approximately 4.05 grams per liter.

Key concepts

Density Calculation

Density is a measure of how much mass is contained in a given volume. To calculate the density of chloroform vapor, the equation we use is:

• Density, \(d = \frac{m}{V} = \frac{MP}{RT}\)

Here, \(m\) is mass, \(V\) is volume, \(M\) is molar mass, \(P\) is pressure, \(R\) is the ideal gas constant, and \(T\) is temperature. The density is expressed in grams per liter (g/L) in this scenario. By using the ideal gas law and substituting the relevant known values into the rearranged equation, we can calculate the density of chloroform vapor with ease. Through this rearrangement, complex relationships between pressure, temperature, volume, and moles simplify into a reliable formula for calculating density in gaseous forms.

Molar Mass

Molar mass is a crucial factor that tells us how much one mole of a substance weighs. For chloroform, which has the chemical formula \(\text{CHCl}_3\), the molar mass is calculated by adding the atomic masses of each atom in the compound:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.01 g/mol
• Chlorine (Cl): 3 atoms each with 35.45 g/mol

The formula to find the molar mass of chloroform becomes:

• \(M = 12.01 + 1.01 + 3 \times 35.45 = 119.37 \, \text{g/mol}\)

This value is used as part of the density formula to link the macroscopic properties of gases to their molecular composition. Understanding molar mass is essential in chemical equations, as it bridges the mass of a sample with the number of moles, allowing wider application of the ideal gas law.

Pressure Conversion

Pressure in scientific calculations can be measured in various units. Common units include atmosphere (atm) and millimeters of mercury (mmHg). When dealing with the Ideal Gas Law, pressure is usually required in atmospheres for calculations involving the gas constant \(R = 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\).To convert the given pressure of chloroform vapor:

• Initial pressure = 797 mmHg
• Conversion = \(\frac{797}{760}\, \text{atm} \approx 1.0487\, \text{atm}\)

This conversion ensures that the pressure is compatible with the units used in the ideal gas equation. Mastery of these conversions helps avoid mistakes in scientific calculations, enabling precise and accurate results.

Temperature Conversion

Temperature is important in calculations involving thermodynamic principles like the ideal gas law. The law's formula requires temperature in Kelvin, since Kelvin is an absolute scale with no negative values.Here’s how to convert Celsius to Kelvin, used in this exercise:

• Starting temperature = 98°C
• Temperature in Kelvin = Celsius + 273.15
• \(T = 98 + 273.15 = 371.15 \, \text{K}\)

This conversion ensures consistency with the gas constant’s unit requirements, allowing for straightforward and correct calculations. The Kelvin scale simplifies calculations where temperature proportionality matters, such as in volume and pressure changes within gases.