Question

A `225*g` mass of glucose is dissolved in a `5*L` volume of water. What is the concentration of the solution? Do we have to account for volume change upon dissolution of the solute?

Answer 1

We must assume that the volume is constant. And I doubt even the dissolution of such a quantity of glucose would cause significant volume change.

Explanation:

`"Molarity"="Moles of solute"/"Volume of solution"`.

And thus,.. `((225*g)/(180.16*g*mol^-1))/(5*L)=0.250*mol*L^-1.`

Note that the way we do the problem, by including the dimensions, is a good check on our calulations. It is all to easy to divide instead of multiply, and vice versa.

Answer 2

`Molarity = 0.25molL^-1 color(white)(x)"or" color(white)(x)0.25moldm^3`

Explanation:

Parameters

Mass of Solute `=225g`

Volume of Solution `=5L`

Recall `->` `Molarity =n/v = "no of moles(mols)"/"volume of solution(L) (dm^3)"`

Also `->` `"no of moles(mols)" = ("Mass of Solute" (C_6H_12O_6))/("Molar Mass of" color(white)(x)C_6H_12O_6)`

`C = 12.0,color(white)(xxx) H = 1.0, color(white)(xxx)O = 16.0`

`"Molar Mass of" color(white)(x)C_6H_12O_6 = color(blue)12.0 xx 6 + color(blue)1.0 xx 12 + color(blue)16.0 xx 6`

`"Molar Mass of" color(white)(x)C_6H_12O_6 = 72 + 12 + 96`

`"Molar Mass of" color(white)(x)C_6H_12O_6 = 180gmol^-1`

`:.` `"no of moles(mols)" = ("Mass of Solute" (C_6H_12O_6))/("Molar Mass of" color(white)(x)C_6H_12O_6)`

`:.` `"n(mols)" = (225g)/(180gmol^-1)`

`color(white)(xxxxxxx)= (225cancelg)/(180cancelgmol^-1)`

`color(white)(xxxxxxx)= 1.25mol`

`rArr` `Molarity =n/v`

`Molarity = (1.25mol)/(5L)`

`Molarity = 0.25molL^-1 color(white)(x)"or" color(white)(x)0.25moldm^3`