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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$^{\large\circ}$ C.

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

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1 Answer

$\pi= iCRT$
$\pi =i\large\frac{n}{V}$$RT$
$x=\large\frac{\pi V}{iRT}$
$\;\;=\large\frac{0.75\times 2.5L}{2.47\times 0.0821Latm K^{-1}mol^{-1}\times 300K}$
$\;\;=0.0308$ mole
Molar mass of $CaCl_2=40+2\times 35.5$
$\Rightarrow 111gmol^{-1}$
$\therefore$ Amount dissolved =$0.0308\times 111$g
$\Rightarrow 3.42g$
answered Aug 6, 2014 by sreemathi.v

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