# 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?

Jul 28, 2017

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:

$\text{Molarity"="Moles of solute"/"Volume of solution}$.

And thus,.. $\frac{\frac{225 \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1}}{5 \cdot L} = 0.250 \cdot m o l \cdot {L}^{-} 1.$

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

$M o l a r i t y = 0.25 m o l {L}^{-} 1 \textcolor{w h i t e}{x} \text{or} \textcolor{w h i t e}{x} 0.25 m o l {\mathrm{dm}}^{3}$

#### Explanation:

Parameters

Mass of Solute $= 225 g$

Volume of Solution $= 5 L$

Recall $\to$ $M o l a r i t y = \frac{n}{v} = \text{no of moles(mols)"/"volume of solution(L) (dm^3)}$

Also $\to$ "no of moles(mols)" = ("Mass of Solute" (C_6H_12O_6))/("Molar Mass of" color(white)(x)C_6H_12O_6)

$C = 12.0 , \textcolor{w h i t e}{\times x} H = 1.0 , \textcolor{w h i t e}{\times x} O = 16.0$

$\text{Molar Mass of} \textcolor{w h i t e}{x} {C}_{6} {H}_{12} {O}_{6} = \textcolor{b l u e}{12.0} \times 6 + \textcolor{b l u e}{1.0} \times 12 + \textcolor{b l u e}{16.0} \times 6$

$\text{Molar Mass of} \textcolor{w h i t e}{x} {C}_{6} {H}_{12} {O}_{6} = 72 + 12 + 96$

$\text{Molar Mass of} \textcolor{w h i t e}{x} {C}_{6} {H}_{12} {O}_{6} = 180 g m o {l}^{-} 1$

$\therefore$ "no of moles(mols)" = ("Mass of Solute" (C_6H_12O_6))/("Molar Mass of" color(white)(x)C_6H_12O_6)

$\therefore$ $\text{n(mols)} = \frac{225 g}{180 g m o {l}^{-} 1}$

$\textcolor{w h i t e}{\times \times \times x} = \frac{225 \cancel{g}}{180 \cancel{g} m o {l}^{-} 1}$

$\textcolor{w h i t e}{\times \times \times x} = 1.25 m o l$

$\Rightarrow$ $M o l a r i t y = \frac{n}{v}$

$M o l a r i t y = \frac{1.25 m o l}{5 L}$

$M o l a r i t y = 0.25 m o l {L}^{-} 1 \textcolor{w h i t e}{x} \text{or} \textcolor{w h i t e}{x} 0.25 m o l {\mathrm{dm}}^{3}$