At 300 K, 36 g of glucose, `C_6H_12O_6` present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the concentration of the other solution.

Explanation

Apply the relation `pi V = cRT` ...(i)
First case
Given that T = 300 K, P = 4.98 bar
Mass of glucose = 36 g, Molecular mass = 180 u
`c = 36/180 = 0.2`
Substituting the values in equation (i), we get
`4.98 xx 1 = 0.2 xx R xx 300 ` ...(ii)
Let the concentration of glucose in the second solution be `c_1` moles per litre.
Substituting the values in equation (i) again, we get

`1.52 xx 1 = c_1 xx R xx 300` ....(iii)
Dividing (ii) by (iii), we have
`(4.98) xx1)/(1.52 xx 1) = (0.2 xx R xx 300)/(c_1 xx R xx 300) `
Simplifying the above equation, we get
`c_1 = (0.2 xx 1.52)/(4.98) = 0.061` moles/litre