Question

2. For the solution of the given substance,
[
begin{array}{c}
Delta T_{b}=100.5-100=0.5^{circ}
Delta T_{b}=K_{b} m
end{array}
]
( therefore )
[
m=frac{Delta T_{b}}{K_{b}}=frac{0.5}{0.50}=1
]
Thus, the given solution is a molal solution, i.e., 1 mole of the solute is present in 1000 g of the solvent. since the solvent is water, its density may be taken to be equal to one. ( therefore ) Volume of solution ( = ) Volume of solvent
[
=frac{1000}{1}=1000 mathrm{mL}=1 mathrm{L}
]
Thus, ( n=1, V=1 L, T=27^{circ} mathrm{C}=300 mathrm{K} ) and ( R=0.0821 )
( mathrm{L} ) atm ( mathrm{K}^{-1} mathrm{mol}^{-1} )
( because V=n R T )
:
( pi=frac{n R T}{V}=frac{1 times 0.0821 times 300}{1} )
[
=24.63 mathrm{atm}
]
Hence, the osmotic pressure of the given solution is 24.63
atm.

# An aqueous solution of a non-volatile and non-electrolytic substance boils at 100.5°C. Calculate the osmotic pressure of this solution at 27°C. Ko for water per 1000 g = 0.50.

Solution