A 1 L solution is prepared by dissolving 0.2 mol NaCl in water at 300K . Find out the value of 2 pi where pi is the osmotic pressure of resulting solution. Assume no change in volume of solution.Use R=0.08L atm mol ^(1-1)K^(-1)

A solution of sucrose (mol. wt. 342 ) is prepared by dissolving 69.3 g of it per litre of the solution. What is the osmotic pressure at 300 K ?

The osmotic pressure M/10 solution of cane sugar (Molar mass 342 g mol-1) at 300 K equal to [R = 0.082 L atm k ^(-1) mol^(-1) ):

When 0.1 M Pb(NO_(3))_(2)(aq) solution is titrated with 0.1 M Kl(aq) solution at 300 K temperature then what will be the osmotic pressure of final solution (in atm) at equivalence point? [Use : R=0.08 L-atm mol^(-1) K^(-1) ]

A solution was prepared by dissolving 6.0 g an organic compound in 100 g of water. Calculate the osmotic pressure of this solution at 298 K ,when the boiling point of the Solution is 100.2^(@)C . ( K_(b) for water = 0.52 K m^(-1) ), R=0.082 L atm K^(_1) mol^(-1) )

The osmotic pressure of 0.2 molar solution of urea at 300 K(R = 0.082) litre atm mol^(-1)K^(-1) is

A solution is prepared by dissolving 0.6 g of urea (molar mass =60 g"mol"^(-1) ) and 1.8 g of glucose (molar mass =180g"mol"^(-1) ) in 100 mL of water at 27^(@)C . The osmotic pressure of the solution is: (R=0.08206L "atm"K^(-1)"mol"^(-1))

Calculate the osmotic pressure of 0.2 M glucose solution at 300 K. (R=8.314 J mol^(-1) K^(-1))

A decinormal solution of NaCl exerts an osmotic pressure of 4.6 atm. at 300 K. Calculate its degree of dissociation. (R=0.082Latm.K^(-1)mol^(-1))

200mL of an aqueus solution of a protien 1.26 g of the protein. The osmotic pressure of such a solution is found to be 2.7xx10^(-3) bar at 300 K. Calculate the molar mass of protein (R=0.083 L bnar mol^(-1)K^(-1))