$\begin{array}{1 1}3.42g\\4.42g\\6.32g\\5.42g\end{array} $

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$\pi= iCRT$

$\pi =i\large\frac{n}{V}$$RT$

$\pi=\large\frac{inRT}{V}$

$x=\large\frac{\pi V}{iRT}$

$\;\;=\large\frac{0.75\times 2.5L}{2.47\times 0.0821Latm K^{-1}mol^{-1}\times 300K}$

$\;\;=0.0308$ mole

Molar mass of $CaCl_2=40+2\times 35.5$

$\Rightarrow 111gmol^{-1}$

$\therefore$ Amount dissolved =$0.0308\times 111$g

$\Rightarrow 3.42g$

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