Question

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

Answer 1

The correct answer is: 3.42 g
We know that,
$π=i\frac{n}{V}RT$⇒π=iw/MVRT
$⇒w=\frac{πMV}{iRT}$
$π=0.75atm$
V=2.5L
i=2.47
T=(27+273)K=300K
Here,
$R = 0.0821\, L\, atm\, K^{-1} mol^{-1 }$
$M = 1 \times 40 + 2 \times 35.5$
$= 111g mol^{-1}$
Therefore, $w=\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}$
= 3.42 g
Hence, the required amount of $CaCl_2$ is 3.42 g.