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Determine the amount of CaCl_2 (i = 2.47) dissolved in 2.5 litre of water such that its osmoticpressure is 0.75 atm at 27^@C . |
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Answer» Solution :For a solute undergoing dissociation, the following EQUATION may be used `pi = ICRT = i n/V RT ` (i is van.t Hoff.s factor) `pi = (pi xx V)/(i xx R xx T)` Substituting the values, we GET `n = (0.75 atm xx 2.5L)/(2.47 xx 0.0821 L atm K^(-1) "mol"^(-1) xx 300 K) = 0.0308` mole Molar mass of `CaCl_2 = 40 + 2 xx 35.5 = 111 g "mol"^(-1)` ` therefore ` Amount dissolved = 0.0308 x 111 g = 3.42 g. |
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