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At 27°C, 36 g of glucose per litre has an osmotic pressure of 4.92 atmosphere. If the osmotic pressure of another solution of glucose is 1.5 atmosphere at the same temperature, what would be its - Chemistry (Theory)

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Question

At 27°C, 36 g of glucose per litre has an osmotic pressure of 4.92 atmosphere. If the osmotic pressure of another solution of glucose is 1.5 atmosphere at the same temperature, what would be its concentration?

Numerical
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Solution

For the first solution of glucose:

Suppose the molecular mass of glucose is M',

π = 4.92 atm, n = `36/M`, V = 1 L, T = 27°C = 300 K

∴ According to van’t Hoff equation nV = nRT, we have

4.92 × 1 = `(36/(M'))` × R × 300    ...(i)

For the second solution of glucose:

π = 1.5 atm, `n = w/(M')`, V = 1 L, T = 300 K

where w is the mass of glucose dissolved per litre of the solution.

∴ `1.5 xx 1 = w/(M') xx R xx 300`    ...(ii)

Dividing eq. (i) by eq. (ii), we have

`4.92/1.5 = 36/w`

or `w = (1.5 xx 36)/4.92`

= 10.98 g

Hence, the concentration of the second solution is 10.98 g L−1.

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Chapter 2: Solutions - NUMERICAL PROBLEMS [Page 120]

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Nootan Chemistry Part 1 and 2 [English] Class 12 ISC
Chapter 2 Solutions
NUMERICAL PROBLEMS | Q 2. | Page 120
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