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?

2 Answers
Jul 28, 2017

Answer:

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.

Jul 28, 2017

Answer:

#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#