Determine the amount (in mg) of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.                (Take RT = 25)

Formula used:

$\mathrm{\pi }=\mathrm{iCRT}=\mathrm{i}\frac{\mathrm{n}}{\mathrm{V}}\mathrm{RT}$

Here, i the van't Hoff factor, C is the molar concentration, R is the ideal gas, T is the temperature, n is the number of moles and V is the volume.

Substitute values in the above equation:

$$\begin{gathered} 0.75=2.47\times \frac{\mathrm{n}}{2.5}\times 0.0821\times 300 \mathrm{K} (273+27°\mathrm{C}) \\ \mathrm{n}=0.0308 \mathrm{mol} \end{gathered}$$

Mass of the solute is calculated as:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}} \\ 0.0308=\frac{\mathrm{mass}}{111 \mathrm{g}/\mathrm{mol}} \\ \mathrm{mass}=3.4210 \mathrm{g}=3412.0 \mathrm{g} \end{gathered}$$