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NCERT Class XII Chemistry
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Question : 40 of 53
Marks: +1, -0
Determine the amount of CaCl2_2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Solution:  
From the expression, π=i nVRT\pi = i \ \frac{n}{V} R T
or, n= π×Vi×R×Tn = \ \frac{\pi \times V}{i \times R \times T}
= 0.75 atm×2.5 L2.47×0.0821 L atm K−1 mol−1×300 K= \ \frac{0.75\ \mathrm{atm} \times 2.5\ \mathrm{L}}{2.47 \times 0.0821\ \mathrm{L}\ \mathrm{atm}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1} \times 300\ \mathrm{K}}
=0.0308 mole=0.0308\ \mathrm{mole}
Molar mass of  CaCl2=40+2×35.5\ \mathrm{CaCl}_2 = 40 + 2 \times 35.5
=111 g mol−1=111\ \mathrm{g}\ \mathrm{mol}^{-1}
∴  \therefore\; Amount of  CaCl2\ \mathrm{CaCl}_2 dissolved =0.0308×111=0.0308 \times 111
=3.42 g=3.42\ \mathrm{g}
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