Question

Calculate the osmotic pressure of $4.5g$ of glucose (Molar mass $=180$) dissolved in $100ml$ of water at $298K$.
(Given : $R=0.0821Latm{mol}^{-1}{k}^{-1}$

Answer 1

Osmotic pressure is proportional to the molarity, $C$, of the solution at a given temperature $T$. Thus
$\pi =CRT$ ...(i)
Where,
$\pi =$ osmotic pressure
$R=$ gas constant $=0.0821{dm}^3$ $atm$ ${mol}^{-1}{K}^{-1}$
$T=$ temperature $=298K$
$C=$ molarity $=\frac{n}{V}=\frac{w}{M\times V}$ ...(ii)
Where,
$n=$ no. of moles
$w=$ Weight of solute glucose $=4.5g$
$M=$ Molecular weight of solute glucose $=180g$
$V=$ Volume of solution $=100ml=0.1L$
Putting the values in the formula (i) and (ii) we get :
$\pi =\frac{wRT}{MV}$
$=\frac{4.5\times 0.0821\times 298}{180\times 0.1}$
$=\frac{110.09}{18}$
$=6.12atm$
The osmotic pressure of the solution is $6.12atm$.