Question:

The volume of water to be added to 100 $cm^3$ of 0.5 $NH_2SO_4$ to get decinormal concentration is

Updated On: Jul 7, 2022
  • 400 $cm^3$
  • 500 $cm^3$
  • 450 $cm^3$
  • 100 $cm^3$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

$N_1 V_1 = N_2V_2$ $0.5\,N \times 100\,cm^3\,=\,0.1\,N\times V_1$ $V_1 = 500\,cm^3$ Water to be added = $(500-100) cm^3 = 400 cm^3$
Was this answer helpful?
0
0

Top Questions on Expressing Concentration of Solutions

View More Questions

Concepts Used:

Concentration of Solutions

It is the amount of solute present in one liter of solution.

Concentration in Parts Per Million - The parts of a component per million parts (106) of the solution.

Mass Percentage - When the concentration is expressed as the percent of one component in the solution by mass it is called mass percentage (w/w).

Volume Percentage - Sometimes we express the concentration as a percent of one component in the solution by volume, it is then called as volume percentage

Mass by Volume Percentage - It is defined as the mass of a solute dissolved per 100mL of the solution.

Molarity - One of the most commonly used methods for expressing the concentrations is molarity. It is the number of moles of solute dissolved in one litre of a solution.

Molality - Molality represents the concentration regarding moles of solute and the mass of solvent.

Normality - It is the number of gram equivalents of solute present in one liter of the solution and it is denoted by N.

Formality - It is the number of gram formula present in one litre of solution.