Question

How would you prepare 1.0 L of an aqueous solution of sucrose(C₁₂H₂₂O₁₁) having an osmotic pressure of 15 atm at a temperature of $\mathbf{22}\mathbf{°}\mathbf{C}$? Sucrose is a nonelectrolyte.

Short answer

When 0.21g of sucrose will be added to an aqueous solution, we can obtain the solution of 1.0 L which will possess osmotic pressure of 15 atm.

Step-by-step solution

Step 1:

According to the question, it is given that:

Volume of solvent (V) = 1.0 L

Osmotic pressure ($\pi$) = 15 atm

Temperature (in K) =

R (constant) = 0.08205 L atm/mol/k

The molecular mass of sucrose (C₁₂H₂₂O₁₁) = $12\times \text{12+22}\times \text{1+11}\times \text{16=342g/mol}$

The osmotic pressure for a solution is given by $\pi =\text{iMRT}$, Here M is the molarity of a solution and i is the Van’t Hoff factor which is 1 for a nonelectrolyte solution.

Molarity of solution

By putting the values, we can determine the molarity of the solution i.e.

$\begin{array}{l}15=\text{1}\times \text{M}\times \text{0.0820}\times \text{295}\\ \text{M}=\frac{15}{\text{0.0820}\times \text{295}}\text{=0.620M}\end{array}$

Mass of solute

From molarity, we can determine the mass of solute that will be needed by the relation:

$\begin{array}{l}\text{Molality=}\frac{\text{Massofsolute}\times \text{1000}}{\text{Molarmassofsolute}\times \text{Volumeofsolvent}}\\ \text{0.620=}\frac{\text{Massofsolute}\times \text{1000}}{\text{342}\times \text{1}}\\ \text{Massofsolute=}\frac{\text{0.620}\times \text{342}}{\text{1000}}\text{=0.21g}\end{array}$