Question:

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002. The molality of the solution is

Updated On: Apr 13, 2024
  • 0.11
  • 0.021
  • 0.004
  • 0.222
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The Correct Option is A

Solution and Explanation

$\frac{P_{0}-P}{P_{0}}=\frac{W_{2}}{M_{2}}\times\frac{M_{1}}{W_{1}}$
$0.002=\frac{W_{2}}{M_{2}}\times\frac{18}{1000}$
$\frac{w_{2}}{M_{2}}=0.11\, mole$
$Molality=\frac{w_{2}}{M_{2}}\times\frac{1000}{W_{1}}=0.11\times\frac{1000}{1000}=0.11 m$
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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.