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?

Answer 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.

Answer 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.