Question

2.65g of S (colloidal solution) in 100ml solution shows Osmotic pressure of 2.463atm at ${{27}^{\circ }}C$.How many S atoms are associated in colloidal sol?(Solution constant = $0.0821atmmo{{l}^{-1}}{{K}^{-1}}$).

Answer 1

For solving this problem we use the Van’t Hoff equation,
$\pi =CRT$, C= Concentration
$C=\dfrac{n}{V}$, n=number of moles and V is the volume of the solution.
Here temperature is used in Kelvin, ${{27}^{\circ }}C$=300K

Complete answer:
So in the question it is given that in 100ml colloidal solution of S, the weight of the S in the solution is 2.65g. The osmotic pressure is given as 2.463atm and the temperature is ${{27}^{\circ }}C$. We have to find the number of S atoms that associate to form the colloidal solution.
For solving the problem, we know the equation for Osmotic pressure, i.e.
$\pi =CRT=\dfrac{nRT}{V}=\dfrac{WRT}{MV}$,
Let’s write the given data in the question and convert all the units in a standard form.
Volume of the colloidal solution, $V= 100ml =0.1L$
Weight of the Sulphur in the solution, $W =2.56g$
Temperature =${{27}^{\circ }}C$ $=300K$
And now let’s find the S atoms associated in the colloidal solution.
The molar mass of the S, M = 32x g/mol, here x is the number of S atoms.
Now substitute all the values in the equation,
$\pi =\dfrac{WRT}{MV}$
$\left( 2.463 \right)=\dfrac{2.56\times 0.0821\times 300}{32\times x\times 0.1}$
Write the equation for x, the equation changes to,
$x=\dfrac{2.56\times 0.0821\times 300}{32\times 0.1\times 2.463}$
$x=8$

So the number of S atoms associated to form the colloidal solution is eight(8).

Note: If in the question concentration was asked then we will solve for C.
All the units should be changed to the standard form before solving the problem.