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?

Options

A. `11 g L^(-1)`
B. `22 g L^(-1)`
C. `36 g L^(-1)`
D. `42 g L^(-1)`

Answer

A

Explanation

`pi V = n_(2) RT`
`pi = W_(2)/(M_(2) xx V) xx R xx T`
`:. 4.98 = (36 xx R xx 300)/(180 xx 1 (L))`
`R= 0.083 "bar L " mol^(-1) K^(-1)`
`pi = CRT`
`C=pi/(RT) = (1.52 "bar")/(0.083 "bar mol"^(-1) K^(-1) xx 300 K)`
`C= 1.52/(0.083 xx 300) molL^(-1) = 10.98 approx 11gL^(-1)`