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Equal weight of a solute are dissolved in equal weight of two solvents \(A\) and \(B\) and formed very dilute solution. The relative lowering of vapour pressure for the solution \(B\) has twice the relative lowering of vapour pressure for the solution \(A\). If \(M_{A}\) and \(M_{B}\) are the molecular weights of solvents \(A\) and \(B\) respectively, then: (a) \(M_{A}=M_{B}\) (b) \(M_{B}=2 \times M_{A}\) (c) \(M_{A}=4 M_{B}\) (d) \(M_{A}=2 M_{B}\)

Short Answer

Expert verified
Option (b) is the correct answer, where \( M_{B}=2 \times M_{A} \).

Step by step solution

01

Understand the Concept of Relative Lowering of Vapor Pressure

Relative lowering of vapor pressure is related to the mole fraction of the solute in the solution and is given by Raoult's law as: \( \frac{p^0 - p}{p^0} = \frac{n_{solute}}{n_{solvent}} \), where \(p^0\) is the vapor pressure of the pure solvent, \(p\) is the vapor pressure of the solution, and \(n_{solute}\) and \(n_{solvent}\) are the number of moles of solute and solvent respectively. Since equal weights of solute are dissolved in equal weights of both solvents, the number of moles of solute is the same in both solutions.
02

Relate Vapor Pressure Lowering to the Molar Masses

Using the relation \( n = \frac{w}{M} \) where \(w\) is the weight and \(M\) is the molar mass, we can deduce the ratio of the relative lowering of vapor pressures to be \( \frac{p^0_B - p_B}{p^0_B} \) is twice \( \frac{p^0_A - p_A}{p^0_A} \) since \( \frac{p^0 - p}{p^0} \) is proportional to \( \frac{n_{solute}}{n_{solvent}} = \frac{w_{solute}/M_{solute}}{w_{solvent}/M_{solvent}} \) and the weight of the solute and weight of the solvents are constant.
03

Set Up the Equation for Molar Mass Ratios

Therefore, the ratio of the molar masses is given by \( \frac{M_B}{M_A} = 2 \), since the relative lowering of vapor pressure for solution B is twice that for solution A and all the other factors cancel out.
04

Determine the Correct Answer

Comparing the ratio that we have found with the provided options, it is clear that \( M_B = 2 \times M_A \), which corresponds to option (b).

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Key Concepts

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

Raoult's Law
Raoult's law is a fundamental principle in chemistry that describes how the vapor pressure of a solution is related to the concentration of solute particles in the solution. Simply put, it states that the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent. This relationship can be mathematically expressed as:
\( P = P^0_{solvent} \times X_{solvent} \),
where \( P \) is the vapor pressure of the solution, \( P^0_{solvent} \) is the vapor pressure of the pure solvent, and \( X_{solvent} \) is the mole fraction of the solvent in the solution. Understanding Raoult's law is essential for solving problems related to the vapor pressure of solutions, including the determination of molar masses and calculation of mole fractions.
Molar Mass Determination
Determining the molar mass is vital for interpreting many aspects of chemical behavior, and it can be done by measuring changes in physical properties, such as vapor pressure. In the provided exercise, we see how the relative lowering of vapor pressure assists in determining the molar mass of the solvents. By analyzing the ratio of the vapor pressure reduction, we indirectly compare molar masses. The use of the formula:
\( n_{solvent} = \frac{w_{solvent}}{M_{solvent}} \)
allows us to calculate the number of moles based on the known weight and the unknown molar mass. When the weight of solute and solvent is constant, as in our exercise, the relative lowering of vapor pressure can be a direct indicator of the molar mass ratio between two solvents.
Mole Fraction
The mole fraction is a dimensionless quantity that represents the ratio of the number of moles of a component to the total number of moles of all components in the solution. It is an essential concept when discussing mixtures and solutions in chemistry. For a two-component system such as a solute dissolved in a solvent, the mole fraction of each component is defined as:
\( X_i = \frac{n_i}{n_{total}} \)
where \( X_i \) is the mole fraction of component \( i \) and \( n_i \) and \( n_{total} \) are the number of moles of component \( i \) and the total number of moles, respectively. In the context of the exercise, Raoult's law connects mole fraction to vapor pressure, illustrating that the mole fraction is a crucial parameter in determining physical properties of solutions such as boiling point, freezing point, and vapor pressure.

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Most popular questions from this chapter

The normal boiling point of water is \(373 \mathrm{~K}\). Vapour pressure of water at temperature \(T\) is 19 \(\mathrm{mm} \mathrm{Hg}\). If enthalpy of vaporisation is \(40.67 \mathrm{~kJ} / \mathrm{mol}\), then temperature \(T\) would be (Use : \(\log 2=0.3, R: 8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ): (a) \(250 \mathrm{~K}\) (b) \(291.4 \mathrm{~K}\) (c) \(230 \mathrm{~K}\) (d) \(290 \mathrm{~K}\)

Assuming complete ionisation, the solution having maximum freezing point will be: (a) \(1 \mathrm{M} \mathrm{CaF}_{2}\) (b) \(1.5 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (c) \(2 \mathrm{M} \mathrm{NaCl}\) (d) \(1 \mathrm{M} \mathrm{AgNO}_{3}\)

The boiling point of an azeotropic mixture of water-ethanol is less than that of both water and ethanol. Then: (a) the mixture will show negative deviation from Raoult's law (b) the mixture will show positive deviation from Raoult's law (c) the mixture will show no deviation from Raoult's law (d) this mixture cannot be considered as true solution

If \(M_{\text {normal }}\) is the normal molecular mass and \(\alpha\) is the degree of ionization of \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), then the abnormal molecular mass of the complex in the solution will be: (a) \(M_{\text {normal }}(1+2 \alpha)^{-1}\) (b) \(M_{\text {normal }}(1+3 \alpha)^{-1}\) (c) \(M_{\text {normal }}(1+\alpha)^{-1}\) (d) equal to \(M_{\text {normal }}\)

\(6.0 \mathrm{~g}\) of urea (molecular weight \(=60\) ) was dissolved in \(9.9\) moles of water. If the vapour pressure of pure water is \(P^{\circ}\), the vapour pressure of solution is: (a) \(0.10 P^{\circ}\) (b) \(1.10 P^{\circ}\) (c) \(0.90 P^{\circ}\) (d) \(0.99 P^{\circ}\)

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