Determine the amount of $CaCl_2$ $(i=2.47)$ dissolved in $2.5 \ L$ of water such that its osmotic pressure is $0.75 \ atm$ at $27^{\circ}C$.

(N/A) We know that the formula for osmotic pressure is $\pi = i \frac{n}{V} RT$,where $n = \frac{w}{M}$.
Thus,$\pi = i \frac{w}{MV} RT$,which rearranges to $w = \frac{\pi MV}{iRT}$.
Given values are:
$\pi = 0.75 \ atm$
$V = 2.5 \ L$
$i = 2.47$
$T = 27 + 273 = 300 \ K$
$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
$M (CaCl_2) = 40 + 2 \times 35.5 = 111 \ g \ mol^{-1}$
Substituting these values into the formula:
$w = \frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}$
$w = \frac{208.125}{60.8361} \approx 3.42 \ g$
Hence,the required amount of $CaCl_2$ is $3.42 \ g$.