Respuesta :
Answer:
Mass percentage of glucose : 22.3%
Mass percentage of sucrose:77.7%
Explanation:
Mass of glucose in sample =
Mass of sucrose in sample = y
Mass of sample = 1.10 g
x + y = 1.10 g ..[1]
Osmotic pressure of the solution = [tex]\pi =3.78 atm[/tex]
Volume of the solution = 25.0 mL = 0.025L ( 1 mL = 0.001 L)
Temperature of the solution =T = 298 K
[tex]\pi=\frac{n_1RT}{V}+\frac{n_2RT}{V}[/tex]
[tex]\pi=(n_1+n_2)\times \frac{RT}{V}[/tex]
[tex]3.78 atm=(n_1+n_2)\times \frac{0.0821 atm L/mol K\times 298 KT}{0.025 L}[/tex]
[tex]0.003862=n_1+n_2[/tex]
[tex]0.003862=\frac{x}{180 g/mol}+\frac{y}{342 g/mol}[/tex]
[tex]0.005556x+0.002924y=0.003862[/tex]..[2]
Solving [1] and [2] we get :
x = 0.2453 g
y = 0.8547 g
Mass percentage of glucose : [tex]\frac{0.2453 g}{1.10 g}\times 100=22.3\%[/tex]
Mass percentage of sucrose: [tex]\frac{0.8547 g}{1.10 g}\times 100=77.7\%[/tex]
