**Question: **What will be the osmotic pressure of a 0.05N solution of sucrose at 6°C? Find the concentration of that solution of glucose. Which would be isotonic of this solution of sucrose.(Molecule weight of sucrose = 342, Molecule weight of glucose = 180, S = 0.0821 litre-atmosphere)

**Solution: **Assuming the molecular weight of sucrose is equal to its equivalent weight. Hence the molar concentration of sucrose solution will be 5M. According to normal solution equation osmotic pressure of sucrose PV = nST –

P = (n/V)ST

= 0.05 x 0.0821 x (273 + 6)

= 1.145 atmosphere

The molar concentrations of isotonic solutions are equal. Therefore, the molar concentration of glucose will also be 0.05M.

Question: A solution of 8.6 gram per litre urea (molecule weight = 60) was isotonic to a 5% solution of an evaporative organic material X. Calculate the molecular weight of X.

Solution: The values of P, S and T are the same in the equation PV = nST for isotonic solutions. Hence their values of n/V are also same.

Molar concentration of urea = concentration in gram per litre/molecule weight

= 8.6/60

There is 5 gram X dissolved in 100 ml of 5% solution of X. So the amount of X in its 1 litre will be 50 gram. Hence concentration in its gram per litre is 50 gram per litre and

Molar concentration of X = molecular weight of 50/molecular weight of X

8.6/60 = 50 / Molecular weight of X

Molecular weight of X = 50 x 60/8.6 = **348.8**