At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Solution and Explanation
Here,
T= 300 K
T = 1.52 bar
\(R = 0.083\, bar\, LK^{-1} mol^{-1}\)
Applying the relation,
\(π = CRT\)
\(⇒C= \frac{π}{RT}\)
\(= \frac{1.52 bar}{(0.083\, bar\,LK^{-1}mol^{-1}\times 300 K)}\)
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.
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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.