Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre

We know that,

$\pi=i \frac{n}{V} \mathrm{R} T$

$\Rightarrow \pi=i \frac{w}{M V} \mathrm{R} T$

$\Rightarrow w=\frac{\pi M V}{i \mathrm{R} T}$

$\pi=0.75 \mathrm{~atm}$

$V=2.5 \mathrm{~L}$

$i=2.47$

$T=(27+273) \mathrm{K}=300 \mathrm{~K}$

Here,

R = 0.0821 L atm K-1mol-1

M = 1 × 40 + 2 × 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 CaCl2 is 3.42 g.