Question:

1L aqueous solution of \(H_2SO_4\) contains 0.02 m mol \(H_2SO_4\). 50% of this solution is diluted with deionized water to give 1L solution (A). In solution (A), 0.01 m mol of \(H_2SO_4\) are added. Total m mols of \(H_2SO_4\) in the final solution is ______ \(× 10^3\) m mols.

Updated On: Dec 21, 2024
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Correct Answer: 0

Solution and Explanation

Initially one litre contains 0.02 mole.
∴ 50% of this solution will contains 0.01 m mol.
After adding 0.01 mol, final solution will contain 0.02 m mol of \(H_2SO_4\).
= 0.02 m mol
= 0.00002 x 103 m molCorrect answer should be 0.

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Concepts Used:

Solutions

A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.

For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.

Types of Solutions:

The solutions can be classified into three types:

  • Solid Solutions - In these solutions, the solvent is in a Solid-state.
  • Liquid Solutions- In these solutions, the solvent is in a Liquid state.
  • Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.

On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:

  1. Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
  2. Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
  3. Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.