At `300 K`, `36 g` of glucose present per litre in its solution has an osmotic pressure of `4.98 bar`. If the osmotic pressure of the solution is `1.52 bar` at the same temperature, what would be its concentration?

Answer

`0.061M`

Explanation

First case
`W_(2)=36g, P=4.98 ba r`,
`Mw_(2)(` glucose `)=180,V=1L,`
`P=(n)/(V)RT`
`4.98=(36)/(180xx1)RT …..(i)`
Second case
`1.52=CRT ….(ii)`
Dividing `Eq. (i)` by `Eq. (ii)`, we get
`(4.98)/(1.52)=(0.2RT)/(CRT)`
`C=(0.2xx1.52)/(4.98)=0.61M`
The concentration of solution `=0.061M`