#### Question

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.

#### Solution

We know that,

`pi = n/V RT`

`=>pi = i w/(MV)RT`

`=>w = (piMV)/iRT`

pi = 0.75 atm

V = 2.5 L

i = 2.47

T = (27+273)K = 300K

Here,

R = 0.0821 L atm K^{-1}mol^{-1}

M = 1 × 40 + 2 × 35.5

= 111g mol^{-1}

Therefore, *w* = `(0.75 x 111 x 2.5)/(2.47xx0.0821xx300)`

= 3.42 g

Hence, the required amount of CaCl_{2} is 3.42 g.

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Solution Determine the amount of CaCl2 dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C Concept: Colligative Properties and Determination of Molar Mass - Osmosis and Osmotic Pressure.