Question #242b5

1 Answer
Jul 17, 2016

Answer:

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#