Home
Class 12
CHEMISTRY
Consider a setup of two urea solution of...

Consider a setup of two urea solution of concentration `C_(1)` and `C_(2)(C_(2)gtC_(1))`, both at temperature `T`, separated by a semi permeable membrane. External pressure `P_(1)` and `P_(2)` respectively are applied on the two solutions. For what values of `P_(1)` and `P_(2)`, osmosis does not occur through the semi permeable membrane? `(R=` universal gas constant)

A

`P_(1)=C_(1)RT` and `P_(2)=C_(2)RT`

B

`P_(1)=` zero and `P_(2)=(C_(2)-C_(1))RT`

C

`P_(1)=C_(2)RT` and `P_(2)=C_(1)RT`

D

`P_(1)=2C_(1)RT` and `P_(1)=(C_(2)+C_(1))RT`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the values of external pressures \( P_1 \) and \( P_2 \) for which osmosis does not occur through a semi-permeable membrane separating two urea solutions of concentrations \( C_1 \) and \( C_2 \) (where \( C_2 > C_1 \)), we can use the concept of osmotic pressure. ### Step-by-Step Solution: 1. **Understanding Osmotic Pressure**: - The osmotic pressure (\( \Pi \)) of a solution can be calculated using the formula: \[ \Pi = C \cdot R \cdot T \] where \( C \) is the concentration of the solution, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. 2. **Setting Up the Conditions**: - We have two solutions: - Solution 1 with concentration \( C_1 \) and external pressure \( P_1 \). - Solution 2 with concentration \( C_2 \) and external pressure \( P_2 \). - For osmosis to not occur, the external pressures must counterbalance the osmotic pressures of the respective solutions. 3. **Equilibrium Condition**: - For no osmosis to occur, the following conditions must be satisfied: \[ P_1 = \Pi_1 \quad \text{and} \quad P_2 = \Pi_2 \] where \( \Pi_1 \) and \( \Pi_2 \) are the osmotic pressures of solutions 1 and 2, respectively. 4. **Applying the Osmotic Pressure Formula**: - For Solution 1: \[ P_1 = C_1 \cdot R \cdot T \] - For Solution 2: \[ P_2 = C_2 \cdot R \cdot T \] 5. **Final Values for Pressures**: - Therefore, the values of \( P_1 \) and \( P_2 \) for which osmosis does not occur are: \[ P_1 = C_1 \cdot R \cdot T \quad \text{and} \quad P_2 = C_2 \cdot R \cdot T \] ### Conclusion: The correct answer is: - \( P_1 = C_1 \cdot R \cdot T \) - \( P_2 = C_2 \cdot R \cdot T \)

To determine the values of external pressures \( P_1 \) and \( P_2 \) for which osmosis does not occur through a semi-permeable membrane separating two urea solutions of concentrations \( C_1 \) and \( C_2 \) (where \( C_2 > C_1 \)), we can use the concept of osmotic pressure. ### Step-by-Step Solution: 1. **Understanding Osmotic Pressure**: - The osmotic pressure (\( \Pi \)) of a solution can be calculated using the formula: \[ \Pi = C \cdot R \cdot T ...
Promotional Banner

Similar Questions

Explore conceptually related problems

V vs T curves at constant pressure P_(1) and P_(2) for an ideal gas are shown below Which is correct ?

Two solutions of KNO_(3) and CH_(3)COOH are prepared separately. The molarity of both is 0.1 M and osmotic pressure is P_(1) and P_(2) , respectively. The correct relationship between the osmotic pressure is

For V versus T curves at constant pressure P_(1) and P_(2) for an ideal gas are shown in the figure given below

Is p = 2 a solution for 2p-1=3 ?

Equal volumes of 0.1 M NaCl and 0.06 M CaCl_(2) solutions are separated by a semi - permeable membrane in container. For this system, choose the correct answer

0.1 M NaCl and 0.05 M BaCl_(2) solutions are separated by a semi-permeable membrane in a container. For this system, choose the correct answer

Figure shows volume - temperature (V-T) graph for same mass of an ideal gas at two different pressures P_1 and P_2 . Hence

For an ideal gas V - T curves at constant pressure P_(1) & P_(2) are shown in figure, from the figure

If P^(@) and P_(s) are vapour pressure of solvent and its solution, respectively, chi_(1) and chi_(2) are mole fractions of solvent and solute, respectively, then

Two cylinder having m_(1)g and m_(2)g of a gas at pressure P_(1) and P_(2) respectively are put in cummunication with each other, temperature remaining constant. The common pressure reached will be