An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because __________
Show Hint
- it gains water due to osmosis
- it loses water due to reverse osmosis
- it gains water due to reverse osmosis
- it loses water due to osmosis
The Correct Option is D
Solution and Explanation
Here's the explanation:
1. Understanding Osmosis:
Osmosis is the natural movement of water molecules from a region of higher water concentration (lower solute concentration) to a region of lower water concentration (higher solute concentration) through a semi-permeable membrane. This process aims to equalize water concentrations on both sides of the membrane.
2. Reverse Osmosis:
Reverse osmosis is the opposite process where water is forced to move from an area of lower water concentration to higher water concentration, requiring the application of external pressure.
3. Analyzing the Mango Scenario:
When an unripe mango is placed in a concentrated salt solution:
- The salt solution has a lower water concentration (higher solute concentration) than the mango's interior
- Water naturally moves out of the mango (higher water concentration) into the salt solution (lower water concentration)
- This water loss causes the mango to shrivel
4. Process Identification:
Since this involves the natural movement of water from high to low concentration areas without external pressure, it is regular osmosis, not reverse osmosis.
Conclusion:
The correct explanation is that the mango loses water due to osmosis.
Final Answer:
The correct answer is (D) it loses water due to osmosis.
Top Questions on Solutions
According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this concept is valid in ecological systems (isotopic mixtures of an element, hydrocarbons mixtures, etc.). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This law is further supported by the fact that Raoult’s law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent.
Answer the following questions:
(a) Give one example of miscible liquid pair which shows negative deviation from Raoult’s law. What is the reason for such deviation?
(b) (i) State Raoult’s law for a solution containing volatile components.
OR
(ii) Raoult’s law is a special case of Henry’s law. Comment.
(c) Write two characteristics of an ideal solution.- A solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water \(K_f\) and \(K_b\) are 1.86 K kg mol\(^{-1}\) and 0.512 K kg mol\(^{-1}\) respectively.
- Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law? Give an example.
- Give reasons:
(a) Cooking is faster in a pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? - In case of association, abnormal molar mass of solute will:
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