Question

\(0.24 \mathrm{~g}\) of a volatile gas upon vaporization gives \(45 \mathrm{~mL}\) vapour at NTP. What will be the vapour density of the substances? (density of \(\mathrm{H}_{2}=1\) ) (a) \(95.39\) (b) \(5.973\) (c) \(95.93\) (d) \(59.73\)

Short answer

The vapour density of the substance is approximately 59.73 (Option d).

Step-by-step solution

Understand the Given Data

The problem provides the following information:- Mass of the gas = 0.24 g- Volume of the gas at NTP = 45 mL- The density of hydrogen, \(\mathrm{H}_2\), is given as 1 for reference. We need to find the vapour density of the gas.

Convert Volume to Standard Molar Volume

At Normal Temperature and Pressure (NTP), 1 mole of an ideal gas occupies 22.4 liters (22,400 mL). We know the vapor occupies 45 mL. First, we will convert this volume to a fraction of the molar volume:\[\frac{45 \, \mathrm{mL}}{22,400 \, \mathrm{mL/mol}} = 0.00200893 \, \text{moles}\].

Calculate the Molar Mass

We know the mass of the vapor is 0.24 g and we have just calculated it as a certain number of moles. To find the molar mass (molecular weight), we use the formula:\[\text{Molar Mass} = \frac{\text{Mass}}{\text{Number of Moles}} = \frac{0.24 \, \text{g}}{0.00200893 \, \text{moles}} \approx 119.48 \, \mathrm{g/mol}\].

Determine Vapour Density

Vapour density is defined as half of the molar mass:\[\text{Vapour Density} = \frac{\text{Molar Mass}}{2} = \frac{119.48}{2} = 59.74\].

Match with Options

From the available options, the vapour density of the volatile gas that we calculated (59.74) closely matches option (d) \(59.73\).

Key concepts

Molar Mass Determination

Molar mass determination is a crucial step in finding out the identity and characteristics of a substance. It refers to the mass of one mole of a substance, essentially the sum of the atomic masses of all the atoms in a molecule. In the context of gases, knowing the molar mass is vital for understanding chemical behavior under various conditions.
To determine the molar mass, you will utilize given data such as the mass of the sample and the number of moles, which can be deduced from the volume it occupies if it's in gas form.

• Step 1: Start with the known mass of the gas.
• Step 2: Calculate the number of moles using the volume provided and the conditions (like NTP, STP).

The formula used for this is: \[\text{Molar Mass} = \frac{\text{Mass of the Sample}}{\text{Number of Moles}}\]
In our exercise, with a mass of 0.24 g and moles being approximately 0.00201, the molar mass is calculated to be approximately 119.48 \( \mathrm{g/mol} \). This gives us a quantitative measurement allowing for further calculations like vapour density.

Ideal Gas Law

The ideal gas law is a central concept in understanding the behavior of gases. It links pressure, volume, temperature, and number of moles of a gas into one comprehensive equation.
The formula is represented as: \[ PV = nRT \] Where:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

This law assumes an ideal situation where gas particles occupy no volume and experience no intermolecular forces. This simplifies calculations and is useful as a comparison model, even though real gases may slightly deviate from this behavior.
In cases like our exercise, which occurs at Normal Temperature and Pressure (NTP), the equation can be simplified, providing an approachable way to connect the physical properties of gas like its volume to learn more about its characteristics.

Vaporization Process

The vaporization process is the transition of a substance from a liquid to a gas. Understanding this process is fundamental in calculating properties like vapour density or molar mass for volatile substances.
Vaporization requires knowledge of the conditions like temperature and pressure under which a substance turns into vapor. Sometimes, this takes place under standard conditions to simplify calculations.

• Volatility: It refers to the tendency of a substance to vaporize. Higher volatility means a substance will convert to vapor more easily.
• Pressure: At higher pressures, vaporization might necessitate higher temperatures. However, at NTP, the molar volume of gas is standardized.

The exercise problem describes vaporization at Normal Temperature and Pressure, typically 273.15 K and 1 atm, making it easier to calculate gas volume which is one mole occupying 22.4 liters.
By comprehending vaporization, students grasp how a liquid or solid changes to a gaseous state, allowing further exploration into gas-related calculations as seen in the exercise.