The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose which is isotonic with blood has the weight-volume $\left(\frac{\mathrm{w}}{\mathrm{v}}\%\right)$ percentage of

Since glucose and blood are isotonic,  ${\mathrm{\pi }}_{\mathrm{glucose}}={\mathrm{\pi }}_{\mathrm{blood}}$

$$\begin{gathered} \mathrm{\pi } =\mathrm{CST} \\ \mathrm{\pi } =\frac{\mathrm{n}}{\mathrm{v}}\mathrm{ST} \\ \mathrm{\pi V} =\mathrm{nST} \end{gathered}$$

 $7.65\times \mathrm{V}=\frac{\mathrm{w}}{180}\times 0.0821\times 310$ [w = amount of glucose present in V litres solution]

 $7.65\times 180=\frac{\mathrm{w}}{\mathrm{V}}\times 0.0821\times 310$

 $\frac{\mathrm{w}}{\mathrm{V}}=\frac{7.65\times 180}{0.0821\times 310}=54.1\mathrm{\hspace{0.17em}}\mathrm{g}\mathrm{\hspace{0.17em}}{\mathrm{L}}^{-1}\hspace{0.17em}\mathrm{or} 5.41\%$

Alternate method :

$$\begin{gathered} \mathrm{\pi }=\mathrm{CST} \\ \mathrm{\pi }=\left(\frac{10\times \left({\displaystyle \frac{\mathrm{w}}{\mathrm{v}}}\%\right)}{\mathrm{MW} \mathrm{of} \mathrm{solute}}\right)\times \mathrm{R}\times \mathrm{T} \\ \frac{\mathrm{w}}{\mathrm{v}}\%=\frac{7.65\times 180}{10\times 0.0821\times 310}=5.41\% ; \end{gathered}$$