At 298 K, 100 ml solution containing 3.002 g of solute gave an osmotic pressure of 2.55 atmospheres. Find the molecular mass of the solute.

Explanation

`T = 298 K`
`v=(100)/(1000)l`
`M_B = ?`
` W_B = 3.002 g`
`pi = 2.55` atm
R = 0.0821 L atm। Kl mol
`pi = (WB)/( M_B.V )RT.`
`M_B =(W_B )/(M_B )`RT
`=(3.002 xx 0.0821 xx 298 )/(2.55 xx (100) /(1000))`
`= 288.2 `g/mol.