the osmotic pressur of glucose soluction is 5.348*10^2Nm^2atm.at 300K.find the amount of glucose in 500ml of soluctiom

Asked by Chaitanya | 26th May, 2017, 07:12: AM

Expert Answer:

M = ∏/iRT
∏ = iMRT
∏ = 5.348 x 102 Nm2 atm
i = 1
R = 0.08206 L atm/mol.K
T = 300 K
 
First let us find the molar concentration in mol/L
M = ∏/iRT
M = 5.348 / 1 x 0.08206 x 300
M = 5.348/24.61 = 0.22 mol/L
M = mol/volume
mol = 0.22 x 0.5 = 0.44
Molar mass of glucose = 180 g/mol
Mass of glucose = 0.44 x 180 = 79.2 g

Answered by Prachi Sawant | 26th May, 2017, 03:36: PM