Amixture of nacl and sucrose (c12h22o11) of combined mass 10.2 g is dissolved in enough water to make up a 250 ml solution. the osmotic pressure of the solution is 7.75 atm at 23°c. calculate the mass percent of nacl in the mixture.

We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:

osmotic pressure = CRT 
C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L
moles of particles = C*V = 0.3189*0.250 =0.0797 mol 
0.0797 = moles of sucrose + 2*moles of salt 

x + 2y = 0.0797 
and 
x(MMsucrose) + y(MMNaCl) = 10.2
342x + 58.5y = 10.2

solve for x and y

x = 0.0252 mol sucrose
y = 0.0273 mol NaCl
 
mass Sucrose = 0.0252(342) = 8.6184 g 
mass NaCl = 0.0273(58.5) = 1.5971 g 
% NaCl = (1.5971 / 10.2)*100 = 15.66%

The answers is bad germs in their