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?

Explanation

Osmotic pressure, `pi=CRT`
First case : Number of moles of glucose, `C_(6)H_(12)O_(6)=(36g)/(180gmol^(-1))=0.2mol`
Mole concentration of solution, `C=0.2/(1L)mol`
`=0.2"mol L"^(-1)`
`4.98"bar"=0.2"mol L"^(-1)xxR(i)xx300k` eqn …(i)
Second case : `1.52=CxxR(ii)xx300K` eqn ...(ii)
Divide eqn (ii) by equation (i)
`(1.52"bar")/(4.98"bar")=(CxxRxx300K)/(0.2molL^(-1)xxRxx300K)=0.305/(1)=C/(0.2molL^(-1))`
`C=0.0610" mol L"^(-1)`