Question

>A 1.10-g sample contains only glucose (C6H12O6) and sucrose (C12H22O11). When a sample is dissolved in water to a total solution volume of 25.0 mL, the osmotic pressure of the solution is 3.78 atm at 298 K.What is the percent mass of the glucose in the sample, and what is the percent mass of sucrose in the sample?

Answer 1

Mass percentage of glucose : 22.3%

Mass percentage of sucrose:77.7%

Step-by-step 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 =
`\pi =3.78 atm`

Volume of the solution = 25.0 mL = 0.025L ( 1 mL = 0.001 L)

Temperature of the solution =T = 298 K

`\pi=(n_1RT)/(V)+(n_2RT)/(V)`

`\pi=(n_1+n_2)* (RT)/(V)`

`3.78 atm=(n_1+n_2)* (0.0821 atm L/mol K* 298 KT)/(0.025 L)`

`0.003862=n_1+n_2`

`0.003862=(x)/(180 g/mol)+(y)/(342 g/mol)`

`0.005556x+0.002924y=0.003862`..[2]

Solving [1] and [2] we get :

x = 0.2453 g

y = 0.8547 g

Mass percentage of glucose :
`(0.2453 g)/(1.10 g)* 100=22.3\%`

Mass percentage of sucrose:
`(0.8547 g)/(1.10 g)* 100=77.7\%`