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Chapter 3: Problem 74

\(0.24 \mathrm{~g}\) 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

Expert verified
The vapor density of the substance is 59.73, so the correct option is (d) 59.73.

Step by step solution

01

Understand the Given Data and Terms

We are given that a volatile gas weighs 0.24 g and occupies 45 mL when vaporized at Normal Temperature and Pressure (NTP). We need to find the vapor density of the gas. Vapor density is defined as half of the molar mass of a gaseous substance. The density of hydrogen is 1, which means its molar mass is approximately 2 g/mol.
02

Convert Volume to Molar Volume at NTP

At NTP, 1 mole of an ideal gas occupies 22.4 L (or 22400 mL). Therefore, to find the number of moles of the gas, we divide the volume of the gas by the molar volume at NTP. The formula is:\[\text{Number of moles} = \frac{\text{Volume at NTP}}{22400 \text{ mL}}\]So, \( n = \frac{45}{22400} \approx 0.0020089 \text{ moles} \).
03

Determine the Molar Mass of the Gas

Using the relationship between mass, moles, and molar mass:\[\text{Molar Mass} = \frac{\text{Mass of the gas}}{\text{Number of moles}}\]we have \( M = \frac{0.24}{0.0020089} \approx 119.45 \text{ g/mol} \).
04

Calculate Vapour Density

Vapour density is defined as half of the molar mass:\[\text{Vapour Density} = \frac{\text{Molar Mass}}{2}\]Thus, \( \text{Vapour Density} = \frac{119.45}{2} = 59.725 \approx 59.73 \).
05

Compare with the Given Options

The calculated vapor density is 59.73. We compare this with the given options and see that option (d) 59.73 is the correct choice.

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Key Concepts

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

Volatile Gas
When you hear the term "volatile gas," think about a gas that easily vaporizes.
Volatility in chemistry refers to the readiness of a substance to transition from a liquid to a gas state. This means such substances have low boiling points and evaporate quickly.

Examples of volatile gases include:
  • Alcohol (ethanol)
  • Ammonia
  • Benzene
For a study of vapor density, it's essential to understand that volatile gases can significantly impact the calculations because they change their state under standard conditions and pressures.
Molar Mass
Molar mass is a fundamental concept in chemistry and describes the mass of a mole of a substance.
It tells us how much one mole of a given substance weighs, usually expressed in grams per mole (g/mol).

For gases, determining molar mass is crucial for understanding characteristics like vapor density. This concept comes into play heavily when predicting how a gas will behave under various conditions.
For example, you can calculate the molar mass by dividing the mass of a sample by the number of moles in the sample:\[\text{Molar Mass} = \frac{\text{Mass of the gas}}{\text{Number of moles}}\]Understanding molar mass helps us comprehend how gases compare to each other in terms of their physical and chemical properties.
Normal Temperature and Pressure (NTP)
Normal Temperature and Pressure, abbreviated as NTP, is a condition used as a standard for scientific measurements.
This standard condition helps in making accurate and consistent comparisons and calculations.

NTP specifies:
  • Temperature: 20掳C (293.15 K)
  • Pressure: 1 atmosphere (101.325 kPa)
Under these conditions, gases behave predictably, allowing chemists to standardize their measurements. When working with gases, knowing the volume occupied by a gas at NTP is critical for calculating its molar volume and, subsequently, other properties like molar mass and vapor density.
Ideal Gas
The ideal gas concept helps in simplifying the study of gases.
An ideal gas follows specific theoretical conditions: it makes certain assumptions about gas particles, such as having no volume and no interactions with each other.

While no real gas is perfectly ideal, this model is useful for:
  • Predicting gas behaviors under various conditions
  • Simplifying equations like the ideal gas law
  • Making calculations involving molar volume at standard conditions
When we talk about gases at NTP or use the ideal gas law, we often make assumptions based on the idea that the gas behaves "ideally." This allows for easier calculation and understanding of gas properties, such as vapor density and molar mass calculations.

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Most popular questions from this chapter

Which one of the following statement is not true about the effect of an increase in temperature on the distribution molecular speeds in a gas? (a) the most probable speed increases (b) the fraction of the molecules with the most probable speed increases (c) the distribution becomes broader (d) the area under the distribution curve remains the same as the under the lower temperature

One litre of gas \(\mathrm{A}\) at \(2 \mathrm{~atm}\) pressure at \(27^{\circ} \mathrm{C}\) and two litres of gas \(\mathrm{B}\) at \(3 \mathrm{~atm}\) pressure at \(127^{\circ} \mathrm{C}\) are mixed in a 4 litre vessel. The temperature of the mixture is maintained at \(327^{\circ} \mathrm{C}\). What is the total pressure of the gaseous mixture? (a) \(3.93 \mathrm{~atm}\) (b) \(3.25 \mathrm{~atm}\) (c) \(4.25 \mathrm{~atm}\) (d) \(6.25 \mathrm{~atm}\)

A gas cylinder has \(370 \mathrm{~g}\) of oxygen at \(298 \mathrm{~K}\) and 30 atm pressure. If the cylinder was heated upto \(348 \mathrm{~K}\) then the valve were held open until the gas pressure was \(1 \mathrm{~atm}\) and the temperature remains \(348 \mathrm{~K}\). What mass of oxygen would escape in this condition? (a) \(349 \mathrm{~g}\) (b) \(359 \mathrm{~g}\) (c) \(329 \mathrm{~g}\) (d) \(339 \mathrm{~g}\)

A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends, the white ammonium chloride ring first formed will be (a) at the centre of the tube (b) near the hydrogen chloride bottle (c) near the ammonia bottle (d) throughout the length of the tube

A general form of equation of state for gases is \(\mathrm{PV}=\) \(\mathrm{RT}\left[\mathrm{A}+\mathrm{B} / \mathrm{V}+\mathrm{C} / \mathrm{V}^{2}+\ldots\right]\), where \(\mathrm{V}\) is the molar volume of the gas and \(\mathrm{A}, \mathrm{B}, \mathrm{C} \ldots .\) are constants for the gas. The values of \(\mathrm{A}\) and \(\mathrm{B}\), if the gas obeys van der Waals equation, are respectively (a) \(1, \frac{a}{27 b}\) (b) \(1, \mathrm{~b}-\frac{a}{\mathrm{RT}}\) (c) \(\frac{a}{R T}, b\) (d) \(\frac{2 \mathrm{a}}{\mathrm{RT}}, \mathrm{b}\)
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