# Question 6ffde

Jun 3, 2016

${M}_{M} = 4.39 \cdot {10}^{4} {\text{g mol}}^{- 1}$

#### Explanation:

Osmotic pressure is simply the pressure that must be applied to a solution in order to prevent the incoming flow of water through a semipermeable membrane. You can derive the equation that gives you the osmotic pressure of a solution that contains a non-electrolyte solute by using the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Isolate $P$ on one side of the equation to get

$P = \frac{n}{V} \cdot R T$

Now, you know that the number of moles of solute per volume of solution gives you the molarity of the solution

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug this into the above equation to get the osmotic pressure, $\Pi$

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

Now, let's say that ${M}_{M} \textcolor{w h i t e}{a} {\text{g mol}}^{- 1}$ is the molar mass of this protein. You can express the number of moles of solute in solution by writing

n = (30.0 color(red)(cancel(color(black)("g"))))/(M_Mcolor(red)(cancel(color(black)("g")))"mol"^(-1)) = 30.0/M_Mcolor(white)(a)"moles"#

You thus have

$\Pi = \frac{30.0}{M} _ M \cdot \frac{1}{V} \cdot R T$

Isolate ${M}_{M}$ on one side of the equation, convert the temperature from degrees Celsius to Kelvin, and plug in your values to get

${M}_{M} = \frac{30.0}{\Pi} \cdot \frac{R T}{V}$

${M}_{M} = \left(30.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(0.0167color(red)(cancel(color(black)("atm")))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(1.00 color(red)(cancel(color(black)("L}}}}\right)$

${M}_{M} = \text{43,919}$

I'll leave the answer rounded to three sig figs and expressed in scientific notation

${M}_{M} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4.39 \cdot {10}^{4} {\text{g mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$