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?

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?

Owen Ambrose · Accepted 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.

#"Molarity"="Moles of solute"/"Volume of solution"#. And thus,.. #((225*g)/(180.16*g*mol^-1))/(5*L)=0.250*mol*L^-1.# As you can see, the inclusion of the dimensions in our solution to the problem serves as a good check on our calculations. It is far too simple to divide rather than multiply, and vice versa.

Owen Ambrose · Answer

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

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#

Xavier Avery · Answer

undefined The concentration of a solution is typically expressed in units of moles per liter (Molarity). To find the concentration of the solution, we first need to calculate the number of moles of glucose present.

Given:
- Mass of glucose = 225 g
- Volume of water = 5 L

The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol.

$$ \text{moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{225 \, \text{g}}{180.16 \, \text{g/mol}} \approx 1.25 \, \text{mol} $$

The concentration of the solution is then calculated by dividing the number of moles of solute by the volume of the solution:

$$ \text{Concentration} = \frac{\text{moles of solute}}{\text{volume of solution}} $$

$$ \text{Concentration} = \frac{1.25 \, \text{mol}}{5 \, \text{L}} = 0.25 \, \text{M} $$

Regarding volume change upon dissolution, for dilute solutions where the volume of the solute is significantly smaller compared to the volume of the solvent, the volume change upon dissolution is usually negligible and can be ignored. However, for highly concentrated solutions or when dealing with solutes that significantly alter the volume upon dissolution (such as gases), it may be necessary to account for volume changes.