Problem 19 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A bottle of cold drink has \(200 \mathrm{ml}\) of liquid in which concentration of \(\mathrm{CO}_{2}\) is \(0.1 \mathrm{M}\). If \(\mathrm{CO}_{2}\), behaves as ideal gas, the volume of \(\mathrm{CO}_{2}^{\circ}\) at \(0^{\circ} \mathrm{C}\) and 1 atm equivalent to the one in cold drink is (a) \(0.2241\) (b) \(0.4481\) (c) \(0.1121\) (d) \(4.481\)

Step-by-Step Solution

Answer

The volume of CO2 at 0°C and 1 atm is 448.1 ml, so the correct option is (b) 0.4481 liters.

Step 1: Calculate moles of CO2

Using the molarity definition, calculate the moles of CO2 present in the 200 ml (0.2 liter) of cold drink solution. Molarity (M) is moles of solute per liter of solution. Thus, moles of CO2 = Molarity × Volume in liters = 0.1 M × 0.2 L = 0.02 moles.

Step 2: Use Ideal Gas Law to find volume at STP

Use the Ideal Gas Law, which is given by PV = nRT, where P is the pressure (1 atm), V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (273.15 K for 0°C). Solve for V: V = (nRT)/P. Plug in the values: V = (0.02 moles × 0.0821 L·atm/mol·K × 273.15 K) / 1 atm.

Step 3: Calculate the volume of CO2 at STP

Calculate the volume by multiplying the constants and the number of moles: V = 0.02 × 0.0821 × 273.15 = 0.4481 liters. Since 1 liter = 1000 ml, V = 0.4481 × 1000 ml = 448.1 ml.

Key Concepts

Molarity and solution concentration

Understanding molarity is critical in chemistry because it helps you describe the concentration of a solution. Molarity is defined as the number of moles of a solute present in one liter of solution. In simpler terms, it tells you how much of a substance is dissolved in a certain volume of liquid.

To calculate molarity, use the formula: \[ Molarity (M) = \frac{moles\text{ of solute}}{liters\text{ of solution}} \]
For our exercise, the molarity is given as 0.1M for \(\mathrm{CO}_2\) in a 200ml (or 0.2 liters) sample of the cold drink. This number means there are 0.02 moles of \(\mathrm{CO}_2\) because \[0.1 \frac{moles}{L} \times 0.2L = 0.02 \text{ moles}\].
Molarity is a foundational concept in stoichiometry, which is the study of quantitative relationships in chemical reactions. It is also invaluable for preparing solutions in the lab, which is a common task in chemistry.

Molar volume of gases at STP

In chemistry, standard temperature and pressure (STP) is a standard set of conditions for measuring the volume of gases: 0°C (273.15 K) and 1 atmosphere of pressure. At STP, one mole of any ideal gas occupies a volume of 22.414 liters, which is known as the molar volume.

However, this exercise requires the calculation of the volume of a non-standard amount of moles at STP. Using the molar volume helps us relate the amount of gas (in moles) to its volume. Knowing that 1 mole of an ideal gas occupies 22.414 liters at STP sets a helpful benchmark for predicting the behavior of a gas under these conditions.

PV=nRT calculations

The Ideal Gas Law, represented as \( PV = nRT \), is a critical equation in chemistry, capturing the relationship between the pressure (P), volume (V), number of moles (n), ideal gas constant (R), and temperature (T) of a given amount of gas.

To solve for the volume of gas like \(\mathrm{CO}_2\) at STP, as in the given exercise, you rearrange the Ideal Gas Law to solve for V: \[ V = \frac{nRT}{P} \].

Here's a closer look at the constants you use in these calculations:

• 'R' is the ideal gas constant, which has a value of 0.0821 L·atm/mol·K.
• 'T' is the temperature in Kelvin, and must be converted from Celsius before use.
• 'P' is the pressure in atmospheres (atm), and at STP it is 1 atm by definition.

By plugging in the values for R, T, and P, and the number of moles of gas, you can calculate the theoretical volume of the gas under those conditions.

Chemistry competitive exam questions

Preparing for chemistry competitive exams requires a firm grasp of various concepts, and practicing problems based on the Ideal Gas Law is essential. This type of question is common because it tests several competencies at once: conceptual understanding of gas laws, stoichiometry, and the ability to perform unit conversions and manipulations of the formula.

In this textbook solution, the steps for solving the Ideal Gas Law problem are broken down, which can aid students in methodically approaching similar problems on an exam. Having a strong foundation in these topics is not only critical for exam success but also forms the bedrock for higher learning in chemistry and related fields. Students should focus on understanding the fundamental principles and practicing a wide variety of problems to excel in their chemistry competitive exams.