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When a liquid that is immiscible with water was steam distilled at \(95.2^{\circ} \mathrm{C}\) at a total pressure of 748 torr, the distillate contained \(1.25 \mathrm{~g}\) of the liquid per gram of water. The vapour pressure of water is 648 torr at \(95.2^{\circ} \mathrm{C}\), what is the molar mass of liquid? (a) \(7.975 \mathrm{~g} / \mathrm{mol}\) (b) \(166 \mathrm{~g} / \mathrm{mol}\) (c) \(145.8 \mathrm{~g} / \mathrm{mol}\) (d) None of these

Short Answer

Expert verified
The molar mass of the liquid is (b) \(166 \text{ g/mol} \).

Step by step solution

01

Understand Dalton's Law of Partial Pressures

According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases. In this steam distillation, the total pressure is caused by the vapor pressures of water and the immiscible liquid combined.
02

Calculate the Partial Pressure of the Liquid

The total pressure in the system is given as 748 torr, and the vapor pressure of water at the given temperature is 648 torr. Therefore, to find the partial pressure of the liquid, subtract the vapor pressure of water from the total pressure: \( P_{liquid} = P_{total} - P_{water} \).
03

Apply Raoult's Law

Considering the distillation is at equilibrium, Raoult's Law can be applied to find the mole fraction of the liquid (\( X_{liquid} \)) using its partial pressure and the given ratio mass of liquid to water. Since water and the liquid are immiscible, their mole fractions can be treated independently. Use the ratio to find the mole fraction: \( X_{liquid} = \frac{Partial~Pressure~of~liquid}{Total~Pressure} \).
04

Calculate Mole Fraction of the Liquid

Following the Raoult's Law, we get the mole fraction of the liquid from the ratio of its pressure to the total pressure: \( X_{liquid} = \frac{P_{liquid}}{P_{total}} \).
05

Calculate Moles of Liquid and Water

Using the given mass ratio (1.25 g of liquid per gram of water) and the mole fractions, write the expression for the moles of liquid and water in the mixture.
06

Determine the Molar Mass of the Liquid

Having the moles of liquid and water from the previous step, calculate the molar mass of the liquid using the mass of the liquid and its moles. The formula is: \( Molar~Mass~of~Liquid = \frac{Mass~of~Liquid}{Moles~of~Liquid} \).
07

Verify the Answer with the Given Options

After finding the molar mass, verify the calculated value against the given options (a), (b), (c), and (d) to determine the correct answer.

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Key Concepts

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

Dalton's Law of Partial Pressures
When we encounter a mixture of non-reactive gases, we can predict the total pressure of the system with Dalton's Law of Partial Pressures. This principle states that the total pressure is simply the sum of the pressures that each gas would exert if it were alone in the container. It becomes particularly useful in steam distillation, where we have water and another liquid that is immiscible with water. Each contributes to the total pressure in the system with their vapor pressures, without interacting with one another. Think of this law as a polite conversation in a room, where each gas speaks its pressure without shouting over the others.

Understanding this law is crucial in steam distillation problems, as it allows us to separate the pressures of each component, making it possible to solve for unknown variables, such as the molar mass of a compound in a steam distillation setup. To apply Dalton's Law effectively, remember to always check for non-reactivity between gases or vapors, as this assumption is foundational to the law's validity.
Raoult's Law
Raoult's Law is a quintessential concept in physical chemistry that helps describe the behavior of liquids in mixtures. To put it simply, Raoult's Law tells us that the partial vapor pressure of a component in a mixture is directly proportional to its mole fraction. This means the more molecules of a component there are in a mixture, the greater its contribution to the total vapor pressure.

The law can be succinctly expressed with the equation: \( P_i = X_i \cdot P_i^\ast \) where \( P_i \) is the partial pressure of component \( i \) in the mixture, \( X_i \) is the mole fraction of \( i \) in the mixture, and \( P_i^\ast \) is the pure component's vapor pressure. It's like assigning a share of the pressure based on how much of each component is present. In the context of distillation, applying Raoult's Law enables us to connect the amount of substance (mole fraction) with a measurable physical property (pressure), which then can lead us to the solution of the molar mass of the unknown liquid.
Vapor Pressure
Imagine you have a liquid in a closed container. Some of its molecules will escape the liquid phase and enter the vapor phase, creating pressure above the liquid. This is what we call vapor pressure, and it's dependent on temperature: the higher the temperature, the more excited the molecules are and the more of them will break free into the gas phase.

Vapor pressure plays a pivotal role in steam distillation. The process consists of heating water and another, immiscible liquid, until their combined vapors reach the total pressure required to boil. The vapor pressures add up due to Dalton's Law, and understanding this can be a key factor in calculating the molar mass of the immiscible liquid, as seen in the exercise. We measure vapor pressure in units like torr, atmospheres, or pascals, and it's essential to keep units consistent when performing calculations.
Mole Fraction
In the world of chemistry, the mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of one substance to the total number of moles of all substances present. The formula for mole fraction \( X_i \) of a substance \( i \) is: \( X_i = \frac{n_i}{n_{total}} \) where \( n_i \) is the moles of substance \( i \) and \( n_{total} \) is the total moles in the mixture.

It's like cutting a pie into slices and you want to know how much of the pie is apple slices versus cherry slices. In the context of distillation and vapor pressure, the mole fraction helps us determine the contribution of each component to the total pressure. Raoult's Law uses mole fraction to relate the partial pressure of a component to its presence in the mixture. When we calculated the mole fraction of the liquid in our exercise using the given mass ratio and Raoult's Law, it was a pivotal step towards finding the molar mass of the immiscible liquid in the steam distillation process.

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Most popular questions from this chapter

One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of \(1.518 \mathrm{~K}\). The degree of association for dimerization of the acid in benzene is \(\left(K_{b}\right.\) for benzene \(\left.=2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right):\) (a) \(60 \%\) (b) \(70 \%\) (c) \(75 \%\) (d) \(80 \%\)

An aqueous solution is \(1.00\) molal in KI. Which change will cause the vapour pressure of the solution to increase ? (a) addition of water (b) addition of \(\mathrm{NaCl}\) (c) addition of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) Addition of \(1.0\) molal KI

If \(1 \mathrm{~m}\) solution of benzoic acid in benzene has a freezing point depression of \(2.4^{\circ} \mathrm{C}\). \(\left(K_{f}=5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\right)\) and boiling point elevation of \(2.53^{\circ} \mathrm{C}\left(K_{b}=2.53^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\right)\) then select the correct statement/s : Statement I : there is dimer formation when undergoing freezing Statement II : there is no change when undergoing boiling Statement III : reverse of I and II Statement IV : dimer formation in freezing and boiling state (a) I, II (b) II, III (c) III, I (d) only I

In a \(0.2\) molal aqueous solution of a weak acid \(H X\) the degree of ionisation is \(0.25\). The freezing point of the solution will be nearest to: \(\left(K_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right.\) ) (a) \(-0.26^{\circ} \mathrm{C}\) (b) \(0.465^{\circ} \mathrm{C}\) (c) \(-0.48^{\circ} \mathrm{C}\) (d) \(-0.465^{\circ} \mathrm{C}\)

The cryoscopic constant of water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\). A \(0.01\) molal acetic acid solution produces a depression of \(0.0194^{\circ} \mathrm{C}\) in the freezing point. The degree of dissociation of acetic acid is: (a) zero (b) \(0.043\) (c) \(0.43\) (d) 1

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