Question

Question #242b5

Answer 1

Here's what I got.

Explanation:

You can't solve this problem without knowing the density of water. When the problem doesn't provide you with this information, you can assume that the density of water is

`rho_"water" = "1 g cm"^(-3)`

This means that your sample of water has a mass of

`1000 color(red)(cancel(color(black)("cm"^3))) * "1 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "1000 g"`

Now, use the molar masses of glucose, `"C"_6"H"_12"O"_6`, and water to calculate how many moles of each substance you have in your solution

`100 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_12"O"_6)/(180.156color(red)(cancel(color(black)("g")))) = "0.5551 moles C"_6"H"_12"O"_6`

`1000color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"`

The mole fraction of a component `i` of a mixture, `chi_i`, is calculated by dividing the number of moles of `i` by the total number of moles present in the mixture

`color(blue)(|bar(ul(color(white)(a/a)chi_i = "number of moles of i"/"total number of moles"color(white)(a/a)|)))`

In your case, the total number of moles present in solution will be

`n_"total" = n_"water" + n_"glucose"`

`n_"total" = "0.5551 moles" + "55.51 moles" = "56.065 moles"`

The mole fraction of glucose, your solute, will be

`chi_"glucose" = (0.5551 color(red)(cancel(color(black)("moles"))))/(56.065color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.01)color(white)(a/a)|)))`

The mole fraction of water, your solvent, will be

`chi_"water" = (55.51color(red)(cancel(color(black)("moles"))))/(56.065color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.99)color(white)(a/a)|)))`

Notice that because your solution only contains glucose and water, you can find the mole fraction of water by

`chi_"water" = 1 - chi_"glucose"`

This once again gets you

`chi_"water" = 1 - 0.1 = 0.99`