Question

Determine the amount of $CaC{l}_2(i=2.4)$ dissolved in $2.5\text{l}$ of water such that its osmotic pressure is $0.75\text{atm}$ at ${27}^{\circ }\text{C}$.
Take $R=\frac{1}{12}\text{latm/molK}$

Answer 1

$\pi =iCRT=i\frac{n}{V}RT$
or, $n=\frac{\pi \times V}{i\times R\times T}=\frac{0.75\times 12\times 2.5}{2.4\times 300}=0.03125\text{mole}$
Molar mass of $CaC{l}_2$ = $40+2\times 35.5=111\text{g/mol}$
$\therefore$ Amount dissolved $=0.03125\times 111$
$=3.47\text{g}$