# Question #6ffde

##### 1 Answer

#### 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

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

Isolate

#P = n/V * RT#

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution")color(white)(a/a)|)))#

Plug this into the above equation to get the osmotic pressure,

#color(blue)(|bar(ul(color(white)(a/a)Pi = c * RTcolor(white)(a/a)|)))#

Now, let's say that **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 = 30.0/M_M * 1/V * RT#

Isolate *degrees Celsius* to *Kelvin*, and plug in your values to get

#M_M = 30.0/Pi * (RT)/V#

#M_M = (30.0 color(red)(cancel(color(black)("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"))))#

#M_M = "43,919"#

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

#M_M = color(green)(|bar(ul(color(white)(a/a)color(black)(4.39 * 10^4 "g mol"^(-1))color(white)(a/a)|)))#