Respostas

2013-10-04T14:08:22-03:00
Sendo as massas molares:
H_2O=18g/mol
Glicose=18og/mol
A fração molar:
x_1=\frac{n_1}{n_1+n_2}
x_1=\frac{x/18}{x/18+x/180}
x_1=\frac{x/18}{11x/180}
x_1=\frac{x}{18}.\frac{180}{11x}
x_1=\frac{180x}{18.11x}
x_1=\frac{10}{11}
logo o da glicose x_2=\frac{1}{11}
1 5 1