An organic substance (M = 169 gram mol–1) is dissolved in 2000 cm3 of water. Its osmotic pressure at 12°C was found to be 0.54 atm. If R = 0.0821 L atm K–1 mol–1, calculate the mass of the solute.
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Solution
Given: T = 273.15 + 12 = 285.15 K
V = 2000 cm3 = 2 L
M2= 169 g mol–1
R = 0.0821 L atm K–1 mol–1
π = 0.54 atm
To find: Mass of solute (W2)
`pi=(W_2RT)/(M2V)`
`W_2=(piM_2V)/(RT)`
`W_2=(0.54xx169xx2)/(0.0821xx285.15)`
`W_2=182.52/23.4108=7.796 g`
Concept: Colligative Properties and Determination of Molar Mass - Introduction
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