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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 1
Here,
T = 300 K
π = 1.52 bar
R = 0.083 bar L K−1 mol−1
Applying the relation,
π = CRT
∴ In the first case, `4.98 = 36/180 xx R xx 300 = 60 R` ...(i)
In the second case, 1.52 = C × R × 300 ...(ii)
Dividing (ii) by (i) we get, C = 0.061 M.
Solution 2
``pi V = w/(M') RT
For the first case,
`4.98 xx 1 = 36/(M') xx R xx T` ...(i)
For the second case,
`1.52 xx 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 = 0.061 M.
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