Question

Question #7deb5

Answer 1

The answer is (c) `"236 g"`

Explanation:

Molality is simply a measure of the number of moles of solute present for every `"1 kg"` of solvent.

This means that a `"1-molal"` glucose solution will contain `1` mole of glucose for every `"1 kg"` of solvent.

`"1 molal " implies " 1 mole glucose" color(white)(.)color(blue)("for every")color(white)(.)"1 kg water"`

As you know, you have

`"1 kg" = 10^3` `"g"`

This means that your solution will contain

`"1 molal " implies " 1 mole glucose" color(white)(.)color(blue)("for every")color(white)(.)10^3color(white)(.)"g water"`

Now, the mass of a solution is always equal to

`"mass solution = mass solute + mass solvent"`

This means that the mass of a `"1-molal"` glucose solution that contains `1` mole of glucose and `10^3` `"g"` of water will be equal to

`"mass 1-molal solution" = "mass of 1 mole glucose" + 10^3color(white)(.)"g"`

To find the mass of `1` mole of glucose, use the compound's molar mass

`1 color(red)(cancel(color(black)("mole glucose"))) * "180.156 g"/(1color(red)(cancel(color(black)("mole glucose")))) = "180.156 g"`

This means that the mass of above solution will be

`"mass 1-molal solution" = "180.156 g" + 10^3color(white)(.)"g"`

`"mass 1-molal solution = 1180.156 g"`

Now, you need your target solution to contain

`"0.2 moles glucose" = "1 mole glucose"/color(red)(5)`

Since solutions are homogeneous mixtures, i.e. they have the same composition throughout, you can say that the mass of solution that will contain `0.2` moles of glucose will be

`"mass solution" = "1180.156 g"/color(red)(5) ~~ color(darkgreen)(ul(color(black)("236 g")))`