Ncert Exercises & Solutions

2.40 Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.

We know that,

$π = i \frac{n}{V} RT$
$\Rightarrow π = i \frac{w}{MV} RT$
$\Rightarrow w = \frac{πMV}{iRT}$
$π = 0.75 atm$
$V = 2.5 L$
$i = 2.47$
$T = (27 + 273) K = 300 K$

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.

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