At 300K ,36g 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

`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`