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The pressure of a mixtures of equal weights of two gases \(X\) and \(Y\) with molecular weight 4 and 40 respectively is \(1.1\) atm. The partial pressure of the gas \(X\) in the mixture is (a) 1 atm (b) \(0.1\) atm (c) \(0.15 \mathrm{~atm}\) (d) \(0.5 \mathrm{~atm}\)

Short Answer

Expert verified
The partial pressure of the gas X in the mixture is 1 atm.

Step by step solution

01

Understanding Dalton's Law of Partial Pressures

Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. The formula is given by: \[ P_{total} = P_X + P_Y \] where \( P_{total} \) is the total pressure of the mixture, and \( P_X \) and \( P_Y \) are the partial pressures of gas X and gas Y respectively.
02

Using the Mole Fraction to Find Partial Pressures

The mole fraction of a gas is the ratio of the number of moles of that gas to the total number of moles of the gases in the mixture. Since the weights of gas X and gas Y are equal and the mole is given by weight divided by molecular weight, their mole ratio can be found as: \[n_X:n_Y = \frac{W_X}{M_X} : \frac{W_Y}{M_Y}\] Substituting the molecular weights we get: \[n_X:n_Y = \frac{W}{4} : \frac{W}{40} = 10:1\] where \(W\) is the weight of each gas which is equal and cancels out.
03

Calculating the Total Moles in the Mixture

The total number of moles in the mixture is simply the sum of the moles of gases X and Y, given by: \[ n_{total} = n_X + n_Y = 10n_Y + n_Y = 11n_Y \]
04

Determining the Mole Fractions

The mole fraction of gas X is given by the number of moles of X divided by the total moles: \[ X_X = \frac{n_X}{n_{total}} = \frac{10n_Y}{11n_Y} = \frac{10}{11} \] Similarly for gas Y: \[ X_Y = \frac{n_Y}{n_{total}} = \frac{n_Y}{11n_Y} = \frac{1}{11} \]
05

Calculating Partial Pressures

Now we apply Dalton's Law to find the partial pressure of gas X: \[ P_X = X_X \times P_{total} \] Substituting the mole fraction and total pressure: \[ P_X = \frac{10}{11} \times 1.1 \,\text{atm} = 1 \,\text{atm} \] Thus, the partial pressure of gas X in the mixture is 1 atm.

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Key Concepts

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

Mole Fraction
The mole fraction is a vital concept in chemistry that helps us understand the composition of a mixture in terms of the proportion of its constituents. It is defined as the ratio of the number of moles of a particular component to the total number of moles of all components in the mixture.

The calculation of mole fraction, typically denoted as \( X \), is straightforward and is given by the formula:
\[ X_i = \frac{n_i}{n_{total}} \]
where:\( n_i \) is the number of moles of the component \( i \), and \( n_{total} \) is the sum of the moles of all components present in the mixture.

Mole fractions are dimensionless quantities and also always add up to one when considering every component in the mixture. They provide an intuitive way to calculate the concentration of a gas in a mixture, preparing us to also explore its partial pressure.
Partial Pressure
When dealing with gas mixtures, the term 'partial pressure' is a key concept. It refers to the pressure that a single gas component in a mixture would exert if it alone occupied the entire volume of the mixture at the same temperature. Dalton's Law of Partial Pressures is fundamental to determining the overall pressure of a gas mixture. It states that the total pressure exerted by a non-reacting mixture of gases is equal to the sum of the partial pressures of each individual gas within the mixture.

The partial pressure of a gas can be determined once the mole fraction is known through the formula:
\[ P_i = X_i \times P_{total} \]
where:\( P_i \) is the partial pressure of the component \( i \), \( X_i \) is the mole fraction of the component, and \( P_{total} \) is the total pressure of the mixture.

Understanding this concept allows students to solve practical problems involving gas mixtures, like calculating how gases will behave under different conditions or how they will contribute to the overall pressure in their environment.
Molecular Weight
Molecular weight is a fundamental chemical property that significantly influences other characteristics of substances, such as the calculation of moles and gas behavior. It is defined as the sum of atomic weights of all the atoms in a molecule and is usually reported in atomic mass units (amu) or grams/mole. The concept is crucial in stoichiometry for converting grams to moles and vice versa, using the formula:
\[ n = \frac{w}{MW} \]
where:\( n \) represents the number of moles, \( w \) is the weight of the substance, and \( MW \) is the molecular weight of the substance.

The role of molecular weight is particularly important in gas laws involving mole fractions and partial pressures since these all involve inter-relational calculations based on the number of moles, which, in turn, are derived from the molecular weight of the gases involved. Thus, a firm grasp of how to determine and apply molecular weight is essential for students as they explore deeper into chemistry and gas laws.

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Most popular questions from this chapter

A gas that follows Boyle's law, Charles' law and Avogadro's law is called an ideal gas. Under what conditions a real gas behaves as ideal gas? (a) Under low pressure and temperature (b) Under high pressure and temperature (c) Under high pressure and low temperature (d) Under low pressure and high temperature

Two atoms \(X\) and \(Y\) are non-polar and electrically symmetrical. What type of intermolecular forces of attraction can be developed between them? (a) Dipole-induced dipole forces (b) London forces or dispersion forces (c) Dipole-dipole forces (d) No forces of any kind.

The correct value of the gas constant \({ }^{\prime} R\) is close to (a) \(0.082\) litre-atmosphere \(\mathrm{K}\) (b) \(0.082\) litre-atmosphere \(\mathrm{K}^{-1} \mathrm{~mol}^{-1}\) (c) \(0.082\) litre-atmosphere \(^{-1} \mathrm{~K} \mathrm{~mol}^{-1}\) (d) \(0.082\) litre \(^{-1}\) atmosphere \(^{-1} \mathrm{~K} \mathrm{~mol}\).

A graph is plotted between pressure and volume at different temperatures. On the basis of the graph what changes will you observe in the volume if (i) the pressure is increased at constant temperature. (ii) the temperature is decreased at constant pressure. (a) volume increases in both the cases (b) volume decreases in both the cases (c) volume increases in (i) and decreases in (ii) (d) volume decreases in (i) and increases in (ii).

A real gas obeying van der Waals equation will resemble ideal gas, if the (a) constents a and \(b\) both are small (b) \(\boldsymbol{a}\) is arge and \(b\) is small (c) \(a\) is \(\mathrm{small}\) and \(b\) is large (d) constants \(a\) and \(b\) both are large.

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