Question

$1\times {{10}^{-3}}kg$ of urea is present in a 400 cubic centimeter solution at 310 K. Calculate the approximate osmotic pressure of the solution.
(R = 8.31$Jmo{{l}^{-1}}{{K}^{-1}}$, molecular weight of urea = 60)
a. 1 atm
b. 2 atm
c. 4 atm
d. 5 atm

Answer 1

Hint: The answer to this question is based on the calculation of molarity of the solution which is given by the ratio of number of moles of solute to the total volume of the solution in litres that is $M=\dfrac{{{n}_{solute}}}{{{V}_{solution(litres)}}}$

Complete step by step answer:
- First of all, let us understand the concept of osmotic pressure. It is defined as “the hydrostatic pressure produced when a solution is separated from a solvent through a semipermeable membrane”.
Osmotic pressure can be calculated by the formula –
\[\pi \] = MRT, where,
\[\pi \]is the Osmotic pressure in atmosphere (atm)
M is the Molarity of the solution (mol/L)
R is the Real gas constant (L atm/K mol)
T is the Temperature (K)
We need to calculate the value of molarity.
- From the formula of molarity which says that molarity is the ratio of number of moles of solute to the total volume of the solution in litres, we can write as,
$M=\dfrac{{{n}_{solute}}}{{{V}_{solution(litres)}}}$
As we can see, we need to calculate the number of moles.
Number of moles =
Weight of urea =$1\times {{10}^{-3}}kg$= 1 g (given)
Molecular weight of urea = 60 g (given)
So, Number of moles =
Volume = 400 cubic centimeters = 0.4 L
Therefore, Molarity =$M=\dfrac{{1}/{60}\;}{0.4}=0.0416M$
Now, we can calculate Osmotic pressure by putting the values.
Therefore,
= MRT = $0.0416\times 0.0821\times 310$ = 1 atm.
So the correct answer is “A”:

Additional Information: Osmotic pressure is a colligative property. Colligative properties are those which depend on concentration of solute.

Note: Note that the values of the gas constant can be written in various forms accordingly to which we need as per the data. The gas constant value in terms of bar will be 8.314$\times {{10}^{-2}}$L. bar. ${{K}^{-1}}mo{{l}^{-1}}$ and in terms of torr can be written as $62.363L.Torr{{K}^{-1}}mo{{l}^{-1}}$