Home
Class 12
CHEMISTRY
Liquid 'M' and liquid 'N' form an ideal ...

Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is : `(x_(M)` = Mole fraction of 'M' in solution, `x_(N)` = Mole fraction of 'N' in solution , `y_(M)` = Mole fraction of 'M' in vapour phase , `y_(N)` = Mole fraction of 'N' in vapour phase)

A

`(x_(M))/(x_(N))=(y_(M))/(y_(N))`

B

`(x_(M)-y_(M))lt(x_(N)-y_(N))`

C

`(x_(M))/(x_(N))lt(y_(M))/(y_(N))`

D

`(x_(M))/(x_(N))gt(y_(M))/(y_(N))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the relationship between the mole fractions of the components in the liquid phase (solution) and the vapor phase, given that the liquids M and N form an ideal solution. ### Step-by-Step Solution: 1. **Identify Given Information:** - Vapor pressure of pure liquid M, \( P^0_M = 450 \, \text{mmHg} \) - Vapor pressure of pure liquid N, \( P^0_N = 700 \, \text{mmHg} \) - Mole fraction of M in solution, \( x_M \) - Mole fraction of N in solution, \( x_N \) - Mole fraction of M in vapor phase, \( y_M \) - Mole fraction of N in vapor phase, \( y_N \) 2. **Use Raoult's Law:** For an ideal solution, Raoult's Law states: \[ y_M = \frac{P^0_M \cdot x_M}{P_{solution}} \] \[ y_N = \frac{P^0_N \cdot x_N}{P_{solution}} \] where \( P_{solution} \) is the total vapor pressure of the solution. 3. **Calculate Total Vapor Pressure:** The total vapor pressure of the solution can be expressed as: \[ P_{solution} = P^0_M \cdot x_M + P^0_N \cdot x_N \] 4. **Express \( y_M \) and \( y_N \):** Substitute \( P_{solution} \) into the equations for \( y_M \) and \( y_N \): \[ y_M = \frac{P^0_M \cdot x_M}{P^0_M \cdot x_M + P^0_N \cdot x_N} \] \[ y_N = \frac{P^0_N \cdot x_N}{P^0_M \cdot x_M + P^0_N \cdot x_N} \] 5. **Form the Ratio of Mole Fractions:** To find the relationship between \( \frac{x_M}{x_N} \) and \( \frac{y_M}{y_N} \), we can divide the equations: \[ \frac{y_M}{y_N} = \frac{P^0_M \cdot x_M}{P^0_N \cdot x_N} \] 6. **Substitute Known Values:** Substitute the values of \( P^0_M \) and \( P^0_N \): \[ \frac{y_M}{y_N} = \frac{450 \cdot x_M}{700 \cdot x_N} \] 7. **Rearranging the Equation:** Rearranging gives: \[ \frac{x_M}{x_N} = \frac{y_M}{y_N} \cdot \frac{700}{450} \] 8. **Conclusion:** Since \( \frac{700}{450} > 1 \), we can conclude: \[ \frac{x_M}{x_N} > \frac{y_M}{y_N} \] This means that the mole fraction of M in the solution is greater than that in the vapor phase, which leads us to the correct statement. ### Final Answer: The correct statement is: \[ \frac{x_M}{x_N} > \frac{y_M}{y_N} \]
Promotional Banner

Topper's Solved these Questions

  • JEE MAINS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise QUESTION|1 Videos
  • JEE MAIN

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise CHEMISTRY|146 Videos
  • JEE MAINS 2020

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise CHEMSITRY|23 Videos

Similar Questions

Explore conceptually related problems

For an binary mixture of A and B with p_(A)^(0)ltp_(B)^(0) : x_(A)= mole fraction of A in solution x_(B)= mole fraction of B in solution y_(A)= mole fraction of A in vapour phase y_(B)= mole fraction of B in vapour phase solution

The vapour pressure of pure liquid M & N are 700mm of Hg and 450mm of Hg respectively. Which of the following option is correct ? Given : X_(N),X_(M)= mole fraction of N & M in liquid phase Y_(N),Y_(M)= mole fraction of N & M in vapour phase

For a ideal liquid solution with P_A^(@)gtP_B^(@) , which relation between X_(A) ((mole fraction of A in liquid phase) and Y_(A) (mole fraction of A in vapour phase) is correct ?

Two liquids A and B form an ideal solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively. Then x is-

The vapour pressures of pure liquids A and B are 400 and 600 mmHg , respectively at 298K . On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are:

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :

The vapour pressure of a certain pure liquid A at 298 K is 40 mbar. When a solution of B is prepared in A at the same temperature, the vapour pressure is found to be 32 mbar. The mole fraction of A in the solution is

The vapour pressures of pure liquids A and B respectively are 600 torr and 500 torr. In a binary solution of A and B , mole fraction of A is 0.25 then total pressure of the solution (in torr) will be