[ 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.
\(0.05 \mathrm{M} \quad 0.05 \mathrm{M} \quad 0.05 \mathrm{M}\)
\(\text { Total } \mathrm{C}_1=0.05+0.05=0.1 \mathrm{M}(\mathrm{NaCl})\)
\(\mathrm{C}_2=0.2 \mathrm{M}(\text { glucose })\)
\(\pi=\left(\mathrm{C}_2-\mathrm{C}_1\right) \mathrm{RT}\)
\(=(0.2-0.1) \times 0.083 \times 300\)
\(=2.49 \text { bar }\)
\(=24.9 \times 10^{-1} \text { bar }\)