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Question
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
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Solution
Given: Temperature (T) = 300 K
R = 0.083 bar L K−1 mol−1
π V = `w/(M') RT`
For the first case,
Osmotic pressure for the first case (π) = 4.98 bar
4.98 × 1 = `36/(M') xx R xx T` ...(i)
For the second case,
Osmotic pressure for the second case (π) = 1.52 bar
1.52 × 1 = `w/(M') xx R xx T` ...(ii)
Dividing eq. (i) by eq. (ii), we get,
`4.98/1.52 = 36/w`
or `w = (1.52 xx 36)/4.98`
= 10.99 g
∴ Moles of solute present in 1 L of solution = `10.99/180`
= 0.061
Hence, the molarity of the solution is 0.061 M.
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