# Question 2fed5

Jun 18, 2017

The mixture contains 11.8 % $\text{NaCl}$ by mass.

#### Explanation:

Let $x = \text{mass of NaCl}$

Then $\left(10.2 - x\right) = \text{mass of sucrose}$.

The formula for osmotic pressure $\Pi$ is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \Pi = i c R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

$i =$ the van't Hoff $i$ factor
$c =$ the molar concentration of a component
$R =$ the universal gas constant
$T =$ the kelvin temperature

We have two solutes, $\text{NaCl}$ and sucrose.

For $\text{NaCl}$:

$i = 2$

$\text{Moles of NaCl" = x/58.44color(white)(l) "mol" = "0.017 11"x color(white)(l)"mol}$

c = ("0.017 11"color(white)(l) "mol")/"0.250 L" = "0.068 45"x color(white)(l)"mol/L"

$T = \text{(23 + 273.15) K = 296.15 K}$

Pi_text(NaCl) = 2 × "0.068 45"x color(red)(cancel(color(black)("mol·L"^"-1"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 296.15 color(red)(cancel(color(black)("K"))) = 3.327xcolor(white)(l) "atm"

For sucrose:

$i = 1$

$\text{Moles of sucrose" = (10.2-x)/342.30 "mol" = ("0.029 80 - 0.002 921"x) color(white)(l)"mol}$

c = (("0.029 80 - 0.002 921"x) color(white)(l)"mol")/"0.250 L" = ("0.1192 - 0.011 69"x) color(white)(l)"mol/L"

Pi_text(sucrose) = 1 × ("0.1192 - 0.011 69"x) color(red)(cancel(color(black)("mol·L"^"-1"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 296.15 color(red)(cancel(color(black)("K"))) = 24.30("0.1192 - 0.011 69"x) color(white)(l)"atm" = (2.897 - 0.2841x) color(white)(l)"atm"

${\Pi}_{\textrm{N a C l}} + {\Pi}_{\textrm{s u c r o s e}} = {\Pi}_{\textrm{\to t a l}}$

$3.327 x \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) + (2.897 - 0.2841x) color(red)(cancel(color(black)("atm"))) = 6.55 color(red)(cancel(color(black)("atm}}}}$

$3.327 x + 2.897 - 0.2841 x = 6.55$

$3.043 x = 3.653$

$x = \frac{3.653}{3.043} = 1.20$

Mass of $\text{NaCl}$ = 1.20 g.

"Mass % NaCl" = (1.20 color(red)(cancel(color(black)("g"))))/(10.2 color(red)(cancel(color(black)("g")))) × 100 % = 11.8 %#