We know that,

π = i \(\frac{n}{V}\)RT

⇒ π = i \(\frac{w}{MV}\)RT

⇒ w = i \(\frac{\pi MV}{iRT}\)

π = 0.75 atm

V = 2.5 L

i = 2.47

T = (27 + 273)K = 300K

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.