Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

\(0.30 \mathrm{~g}\) of a volatile liquid displaces \(90.0 \mathrm{~cm}^{3}\) of air at STP in the Victor Meyer's method. The molecular mass of the liquid is (a) \(54.44 \mathrm{~g}\) (b) \(34.64 \mathrm{~g}\) (c) \(64.76 \mathrm{~g}\) (d) \(74.66 \mathrm{~g}\)

Step-by-Step Solution

Answer

The molecular mass of the liquid is 74.66 g (option d).

Step 1: List Given Information

Start by identifying the given values from the problem statement: the mass of the volatile liquid is \(0.30 \text{ g}\) and the volume of air displaced is \(90.0 \text{ cm}^3\). The problem occurs at Standard Temperature and Pressure (STP).

Step 2: Convert Volume to moles

At STP, 1 mole of an ideal gas occupies \(22,400 \text{ cm}^3\). Calculate the number of moles displaced using the formula: \( \text{moles} = \frac{\text{Volume displaced}}{22,400 \text{ cm}^3}\). Substitute the known volume: \( \frac{90.0}{22,400} \approx 0.00402 \text{ moles}\).

Step 3: Calculate Molecular Mass

The molecular mass can be found using the formula: \( \text{Molecular mass} = \frac{\text{mass of gas}}{\text{moles of gas}}\). Substitute the known values: \( \frac{0.30 \text{ g}}{0.00402 \text{ moles}} = 74.63 \text{ g/mol}\).

Step 4: Choose Closest Option

Compare the calculated molecular mass to the provided choices: (a) 54.44 g, (b) 34.64 g, (c) 64.76 g, (d) 74.66 g. The closest value is (d) 74.66 g.

Key Concepts

Molecular Mass Calculation

Molecular mass, also known as molecular weight, is an important concept in chemistry that helps us understand the amount of matter contained in molecules. To calculate molecular mass, we need to divide the mass of a substance by the number of moles it comprises. This gives us the mass of one mole of the substance, expressed in grams per mole (g/mol).

In the context of Victor Meyer's method used in the exercise, we begin with a volatile liquid. By determining the volume it displaces at standard conditions, we can estimate the number of moles. Once the number of moles is known, we use the formula: \[ \text{Molecular mass} = \frac{\text{mass of gas}}{\text{moles of gas}} \]This allows us to calculate its molecular mass accurately.

Knowing the molecular mass is crucial because it helps in predicting how substances will react with one another. The greater the molecular mass, the heavier the molecule, which might influence its properties and behavior in chemical reactions.

Ideal Gas Law

The ideal gas law is a fundamental principle in chemistry that helps you understand how gases behave under different conditions. It's expressed through the equation \[ PV = nRT \]where:

• \(P\) is pressure,
• \(V\) is volume,
• \(n\) is the number of moles,
• \(R\) is the gas constant, and
• \(T\) is temperature.

This equation allows you to solve for any of the variables if the others are known, making it an incredibly versatile tool for calculations involving gases. For example, it can help determine the volume a gas will occupy at a given pressure and temperature.

However, it's important to remember that the ideal gas law makes some assumptions: all particles are in constant motion and there are no intermolecular forces between gas particles. Real gases do behave a bit differently, particularly under high pressure or low temperature, but this law gives a good approximation for many problems.

The exercise utilizes this concept by assuming that at STP, 1 mole of a gas occupies a standard volume, which simplifies the conversion between volume and moles.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure, abbreviated as STP, is a set of conditions used as a reference point in ideal gas calculations. Standard temperature is defined as 0 degrees Celsius (273.15 Kelvin), and standard pressure is 1 atmosphere (atm). These conditions provide a baseline so that scientists can communicate and compare gas-related results consistently.

At STP, 1 mole of an ideal gas occupies a volume of 22,400 cubic centimeters (cm³) or 22.4 liters (L). This standardization makes it easier to relate the amount of a gas to its volume, as well as simplifying the calculation of its properties under these conditions.

The significance of STP in the exercise is that it allows us to convert the volume of gas displaced by the liquid into moles directly, without needing to adjust for different temperatures or pressures. This direct conversion simplifies the process of determining molecular mass, highlighting the practical utility of these standardized conditions.