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 27⁰C.

Solution

Answer;

π = \frac{inRT}{V} π = \frac{iWRT}{MV} where n = w / m w = \frac{πMV}{iRT}

CaCl₂ → Ca²⁺ + 2Cl^{–

$π = 0.75 atm$}

T = 273 + 27 = 300 K,
M = 1 x 40 + 2x 35.5 = 111g mol^-1
V = 2.5 L
i = 2.47
osmatic pressure,

π = \frac{i n_{2} RT}{V}

w = \frac{π M V}{i R T} t h e r e f o r e w = \frac{0.75 \times 111 \times 2.5}{2.47 \times 0.082 \times 300} = 3.42 g

amount of calcium chloride dissolved is 3.42g

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