what would be osmotic pressure of 0.05 N sucrose solution at 5 degree Celsius? Find out concentration of glucose solution which is isotonic with this sucrose solution.
ans = 9g/L

Dear Student,

For sucrose , molarity = normality since valency=1.

So, 0.05 N = 0.05 M

$$\begin{gathered} T = temperature = 5{ }^{o}C=278 K \\ R = Gas cons\tan t = 0.08206 L\hspace{0.17em}atm mo{l}^{-1} K^{-1} \\ C = concentration= 0.05 M \\ \pi = osmostic pressure \\ Now, \\ \pi = CRT = 0.05 \times 0.08206 \times 278 = 1.1406 atm \end{gathered}$$

Thus, the osmotic pressure is 1.14 atm.

Now, we know that the number of moles of sucrose in 1 L of solution is 0.05.

For two solutions to be isotonic, they should have the same number of moles.
Molar mass of glucose = 180 g mol^-1
Therefore, weight of glucose = 180 x 0.05 = 9 g
For 1 L of solution, concentration of glucose = $\frac{weight}{volume}=\frac{9}{1}=9 g L^{-1}$

Hope it helps.

Regards