Discussion Forum

The osmotic pressure of a solution containing 34.2 g of cane sugar (molar mass =342 g /mol)  in 1L of solution at 20° C is (Given R=0.082 L atm K^-1 mol^-1 )

Answer:

Explanation:

Osmotic pressure, $\pi$= CRT

where, C= concentration of solution

Given, w= 34.3 g, V=1L , |M= 342 g/mol

 T= 20°  C= 20+273=293°  K.

 $\therefore$    $\pi=\frac{W}{MV}RT=\frac{34.2 g}{342 g mol^-1}$

                                                     $\times \frac{0.082 L atmK^{-1}mol^{-1}\times293 K}{1L}$

     =2.40 atm