At 298 K the vapour pressure of pure water is 23.75 mm Hg.
a) At the same temperature calculate the vapour pressure over 10% aqueous solution of an organic compound whose molecular weight is 60 g/mol.
b) What will be the osmotic pressure of 500 ml of this solution at 298 K? [R = 0.0821 L atm/K mol]

Dear Student,

Please find below the solution to the asked query:

a)
Mass of solvent (W_A) = 100 - 10 = 90 g
Mass of Solute (W_B) = 10 g
Molecular mass of the solvent (M_A) = 18 g mol^-1​ 
Molecular mass of the solute (M_B) =  60 g mol^-1​ 
Vapour pressure of pure water (p_A^o) = 23.75 mm Hg
Vapour pressure of Solution (p_A ) = ?

Now, $$\begin{gathered} \left(\frac{p_A^o}{p_A^o-p_A}\right)=\frac{W_A\times M_B}{M_A\times W_B} \\ \left(\frac{23.75}{23.75-p_A}\right)=\frac{90 g\times 60 g mo{l}^{-1}}{18 g mo{l}^{-1}\times 10 g} \\ \left(\frac{23.75}{23.75-p_A}\right) = 30 \\ 23.75 = 30 \left(23.75-p_A\right) \\ {\mathit{p}}_{\mathbf{A}}\mathbf{=}\mathbf{22}\mathbf{.}\mathbf{96}\mathbf{ }\mathit{m}\mathit{m}\mathit{H}\mathit{g} \end{gathered}$$


b) Osmotic pressure:

$\mathrm{\pi }=\frac{{\mathrm{W}}_{\mathrm{B}} \mathrm{R} \mathrm{T}}{{\mathrm{M}}_{\mathrm{B}} \mathrm{V}}=\frac{10 \mathrm{g}\times 0.0821 \mathrm{L} \mathrm{atm} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}\times 298 \mathrm{K}}{60 \mathrm{g} {\mathrm{mol}}^{-1}\times 0.5 \mathrm{L}}=\mathbf{8}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{atm}$


Hope this information will clear your doubts about Solutions.
If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.

Best Wishes !