[ આપેલ : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ]
$\mathrm{NaCl}$ નું સંપૂર્ણ વિયોજન થાય છે તે ધારી લો.
$0.05 \mathrm{M} \quad 0.05 \mathrm{M} \quad 0.05 \mathrm{M}$
$\text { Total } \mathrm{C}_1=0.05+0.05=0.1 \mathrm{M}(\mathrm{NaCl})$
$\mathrm{C}_2=0.2 \mathrm{M}(\text { glucose })$
$\pi=\left(\mathrm{C}_2-\mathrm{C}_1\right) \mathrm{RT}$
$=(0.2-0.1) \times 0.083 \times 300$
$=2.49 \text { bar }$
$=24.9 \times 10^{-1} \text { bar }$