# What are the mass fraction and the molar concentration of glucose (M_text(r) = 180.16) in 320 g of a solution that contains 40 g of glucose?

## $\text{Density of solution = 1400 kg·m"^"-3}$

Jul 4, 2017

The mass fraction of glucose is 0.125. The molar concentration is ${\text{0.97 mol/dm}}^{3}$.

#### Explanation:

Mass Fraction

The mass fraction ${w}_{i}$ of a component $i$ in a mixture is the mass of the component ${m}_{i}$ divided by the total mass ${m}_{\textrm{\to t}}$:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {w}_{i} = {m}_{i} / {m}_{\textrm{\to t}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

In this problem,

${m}_{\textrm{g l u \cos e}} = \text{40 g}$
${m}_{\textrm{\to t}} = \text{40 g + 280 g = 320 g}$

${w}_{\textrm{g l u \cos e}} = \left(40 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))))/(320 color(red)(cancel(color(black)("g}}}}\right) = 0.125$

Molar concentration

The formula for molar concentration $c$ is

color(blue)(bar(ul(|color(white)(a/a)c = "moles"/"cubic decimetres" = n/Vcolor(white)(a/a)|)))" "

n = 40 color(red)(cancel(color(black)("g"))) × "1 mol"/(180.16 color(red)(cancel(color(black)("g")))) = "0.222 mol"

V = 320 × 10^"-3" color(red)(cancel(color(black)("kg"))) × ("1 m"^3)/(1400 color(red)(cancel(color(black)("kg")))) = 2.286×10^"-4" color(white)(l)"m"^3 = "0.2286 dm"^3

$c = {\text{0.222 mol"/"0.2286 dm"^3 = "0.97 mol/dm}}^{3}$