Question

Equal weights of ethane and hydrogen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction to total pressure exerted by hydrogen is (a) \(1: 2\) (b) \(1: 1\) (c) \(1: 16\) (d) \(15: 16\)

Short answer

(d) The fraction of the total pressure exerted by hydrogen is \(15: 16\).

Step-by-step solution

Understand the Problem

We are given that equal masses of ethane (C_2H_6) and hydrogen (H_2) are mixed, and we need to find the fraction of the total pressure that is exerted by hydrogen. The options involve ratios of pressure fractions.

Establish Molar Masses

First, calculate the molar mass of ethane (C_2H_6) and hydrogen (H_2). For ethane, the molar mass is approximately 30 g/mol (C:12, 6H:6), and for hydrogen, the molar mass is 2 g/mol (2H:2).

Determine Moles of Each Gas

Using the definition of moles, from equal masses we can equate:\(\text{mass of } C_2H_6 = \text{mass of } H_2\).By dividing by the molar mass, the number of moles:\\( n_{C_2H_6} = \frac{m}{30} \) and \\( n_{H_2} = \frac{m}{2} \).

Calculate Mole Fraction and Pressure Contribution

The mole fraction is found by dividing the moles of hydrogen by the total moles:\[X_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{C_2H_6}} = \frac{\frac{m}{2}}{\frac{m}{2} + \frac{m}{30}} = \frac{15}{16}\]The pressure exerted is proportional to its mole fraction, thus, hydrogen's pressure contribution is \(\frac{15}{16}\).

Choose the Correct Answer

Based on the calculations, the fraction of the total pressure exerted by hydrogen is \(\frac{15}{16}\). Therefore, the correct answer is option (d) \(15: 16\).

Key concepts

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture of gases. It is defined as the number of moles of a particular substance divided by the total number of moles of all substances present. The formula for mole fraction (X) is:\[ X_A = \frac{n_A}{n_A + n_B + n_C + \ldots} \]Where:

• \(X_A\): the mole fraction of component \(A\)
• \(n_A\): number of moles of component \(A\)
• \(n_B, n_C, \ldots\): number of moles of other components in the mixture

In our scenario, we have hydrogen and ethane, so the mole fraction of hydrogen is calculated by:\[ X_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{C_2H_6}} \]Calculating correctly ensures you understand how each gas contributes to the total mixture concentration. This concept helps us find the pressure exerted by hydrogen in the container.

Molar Mass

Molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). It is an essential concept for converting between the mass of a substance and the number of moles, which is crucial for understanding chemical quantities in reactions and mixtures.To find the molar mass of a compound, you simply add up the atomic masses of all the atoms present. For our example:

• Ethane (\(C_2H_6\)): Two carbon atoms (each with an atomic mass of 12) and six hydrogen atoms (each with an atomic mass of 1) gives \( 2 \times 12 + 6 \times 1 = 30 \, \text{g/mol} \).
• Hydrogen (\(H_2\)): Two hydrogen atoms together have a molar mass of \( 2 \times 1 = 2 \, \text{g/mol} \).

Knowing the molar mass allows us to determine how many moles are present in a given mass, helping us calculate properties like the mole fraction and pressure contribution.

Pressure Contribution

In a gas mixture, each component exerts pressure independently as if it occupied the entire volume of the container by itself. This partial pressure is proportional to its mole fraction. Consequently, the total pressure of the gas mixture is the sum of the individual pressures exerted by each component.The concept of pressure contribution arises from Dalton's Law of Partial Pressures, which states: \[ P_{\text{total}} = P_{A} + P_{B} + \ldots \]Where each \( P \) is the partial pressure of a gas, found by multiplying its mole fraction by the total pressure:\[ P_{A} = X_{A} \cdot P_{\text{total}} \]Applying this to our exercise, the pressure of hydrogen in the mixture is:\[ P_{H_2} = X_{H_2} \cdot P_{\text{total}} \]Understanding pressure contribution helps us to not only discover the portion of total pressure a gas contributes but also to solve practical problems involving gas mixtures, like determining the fraction of total pressure exerted by hydrogen in the given exercise.